Any ideas on how I can prove this expression?

Note: Here are at least some few hints which may help to solve this nice identity. In fact it's hardly more than a starter. But hopefully some aspects are nevertheless useful for the interested reader.

Introduction: The following information is provided:

• Definition of partial Bell polynomials $B_{n,k}$:

I state the definition of Bell polynomials according to Comtet's: classic Advanced Combinatorics section 3.3 Bell Polynomials as coefficients of generating functions. The idea is, that a proper representation via generating functions could help to find the solution.

• The convolution product $x \diamond y$ demystified

We will observe, that the convolution product is strongly related with an iterative representation of the generating functions of the Bell numbers. This enables us to transform the stated identity to gain some more insight.

• Representation of the identity with the help of generating functions

In fact its just another representation regrettably without clever simplifications

• Verification of the identity for $n=2,3$

In order to better see what's going on, the identity is also verified for small $n=2,3$. An analysis of these examples could provide some hints how to appropriately transform the generating functions in the general case.

Definition of partial Bell polynomials $B_{n,k}$:

According to Comtet's Advanced Combinatorics section 3.3 Bell Polynomials we define as follows:

Let $\Phi(t,u)$ be the generating function of the (exponential) partial Bell polynomials $B_{n,k}=B_{n,k}(x_1,x_2,\ldots,x_{n-k+1})$ in an infinite number of variables $x_1,x_2,\ldots$ defined by the formal double series expansion:

\begin{align*} \Phi(t,u)&:=exp\left(u\sum_{m\geq 1}x_m\frac{t^m}{m!}\right)=\sum_{n,k\geq 0}B_{n,k}\frac{t^n}{n!}u^k\\ &=1+\sum_{n\geq 1}\frac{t^n}{n!}\left(\sum_{k=1}^{n}u^kB_{n,k}(x_1,x_2,\ldots)\right) \end{align*} or what amounts to the same by the series expansion: \begin{align*} \frac{1}{k!}\left(\sum_{m\geq 1}x_m\frac{t^m}{m!}\right)^k=\sum_{n\geq k}B_{n,k}\frac{t^n}{n!},\qquad k=0,1,2,\ldots\tag{1} \end{align*}

In the following the focus is put on the representation (1).

Let's use the coefficient of operator $[t^n]$ to denote the coefficient $a_n=[t^n]A(t)$ of a formal generating series $A(t)=\sum_{k\geq 0}a_kt^k$.

We observe for $n\geq 0$:

\begin{align*} B_{n,k}=\frac{n!}{k!}[t^n]\left(\sum_{m\geq 1}x_m\frac{t^m}{m!}\right)^k,\qquad k\geq 0\tag{2} \end{align*}

Note: In the following it is sufficient to consider $B_{n,k}$ for $n,k\geq 1$.

The convolution product $x\diamond y$ (somewhat demystified)

The convolution product $x \diamond y$ for sequences $x=(x_j)_{j\geq 1}$ and $y=(y_j)_{j\geq 1}$ is defined according to this link as

\begin{align*} x \diamond y :=\sum_{j=1}^{n-1}\binom{n}{j}x_jy_{n-j}\tag{3} \end{align*}

The polynomial $B_{n,k}$ can be written using the $k$-fold product $$x^{k_\diamond}=(x_n^{k\diamond})_{n\geq 1}:=\underbrace{x\diamond \ldots \diamond x}_{k \text{ factors}}$$ as

\begin{align*} B_{n,k}=\frac{x_n^{k\diamond}}{k!}, \qquad n,k\geq 1\tag{4} \end{align*}

We obtain according to the definition: \begin{align*} x\diamond y&=\left(0,\sum_{j=1}^{1}\binom{2}{j}x_jy_{2-j},\sum_{j=1}^{2}\binom{3}{j}x_jy_{3-j},\sum_{j=1}^{3}\binom{4}{j}x_jy_{4-j},\ldots\right)\\ &=\left(0,2x_1y_1,3x_1y_2+3x_2y_1,4x_1y_3+6x_2y_2+4x_3y_1,\ldots\right)\\ \end{align*} which implies \begin{align*} x^{1\diamond}&=\left(x_1,x_2,x_3,x_4,\ldots\right) &=&1!(B_{1,1},B_{2,1},B_{3,1},B_{4,1}\ldots)\\ x^{2\diamond}&=\left(0,2x_1^2,6x_1x_2,8x_1x_3+6x_2^2,\ldots\right) &=&2!(0,B_{2,2},B_{3,2},B_{4,2},\ldots)\\ x^{3\diamond}&=\left(0,0,6x_1^3,36x_1^2x_2,\ldots\right) &=&3!(0,0,B_{3,3},B_{4,3}\ldots)\\ \end{align*}

Observe, that the multiplication of exponential generating functions $A(x)=\sum_{k\geq 0}a_k\frac{x^k}{k!}$ and $B(x)=\sum_{l\geq 0}b_l\frac{x^l}{l!}$ gives: \begin{align*} A(x)B(x)&=\left(\sum_{k\geq 0}a_k\frac{x^k}{k!}\right)\left(\sum_{l\geq 0}b_l\frac{x^l}{l!}\right)\\ &=\sum_{n\geq 0}\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{a_k}{k!}\frac{b_l}{l!}\right)x^n\\ &=\sum_{n\geq 0}\left(\sum_{k=0}^n\binom{n}{k}a_kb_{n-k}\right)\frac{x^n}{n!} \end{align*}

According to the definition of $B_{n,k}$ the sequences $x^{k_\diamond}$ generated via the convolution product are simply the coefficients of the vertical generating functions for the Bell polynomials $B_{n,k}, n\geq 1$:

\begin{align*} \frac{1}{1!}\sum_{m\geq 1}x^{1_\diamond}_n\frac{t^n}{n!}&= \frac{1}{1!}\left(\sum_{m\geq 1}x_m\frac{t^m}{m!}\right)=\sum_{m\geq 1}B_{n,1}\frac{t^n}{n!}\\ \frac{1}{2!}\sum_{m\geq 1}x^{2_\diamond}_n\frac{t^n}{n!}&= \frac{1}{2!}\left(\sum_{m\geq 1}x_m\frac{t^m}{m!}\right)^2=\sum_{m\geq 2}B_{n,2}\frac{t^n}{n!}\\ &\qquad\qquad\cdots\\ \frac{1}{k!}\sum_{m\geq 1}x^{k_\diamond}_n\frac{t^n}{n!}&= \frac{1}{k!}\left(\sum_{m\geq 1}x_m\frac{t^m}{m!}\right)^k=\sum_{m\geq k}B_{n,k}\frac{t^n}{n!}\\ \end{align*}

We observe: A convolution with $x^{1\diamond}$ corresponds essentially to a multiplication of the generating function $\sum_{m\geq 1}x_m\frac{t^m}{m!}$

In order to keep complex expressions better manageable, we introduce some abbreviations:

$$B_{n,k}^{f}(x) := B_{n,k}(f^\prime(x),f^{\prime\prime}(x),\ldots,f^{(n-k+1)}(x))$$

The $n$-th derivatives will be abbreviated as

$$f_n:=\frac{d^n}{dx^n}f(x)\qquad\text{ and }\qquad g_n:= \frac{d^n}{dx^n}g(x),\qquad n \geq 1$$

we also use OPs shorthand $a=(f_1,f_2,\ldots)$ and $b=(g_1,g_2,\ldots)$.

According to the statements above the expression

$$B_{n,k}\left(f^\prime(x),f^{\prime\prime}(x),\ldots,f^{(n-k+1)}\right)_{(f\rightarrow g)^c} =\frac{\left(a^{(k-c)_{\diamond}}\diamond b^{c_\diamond}\right)_n}{(k-c)!c!}$$

can now be written as coefficients of the product of the generating functions

\begin{align*} \sum_{n\geq k}&B^f_{{n,k}_{(f\rightarrow g)^c}}\frac{t^n}{n!} =\frac{1}{(k-c)!}\left(\sum_{m\geq1}f_m\frac{t^m}{m!}\right)^{k-c} \frac{1}{c!}\left(\sum_{m\geq1}g_m\frac{t^m}{m!}\right)^{c} \end{align*}

Representation of the identity via generating functions

We are now in a state to represent OPs identity with the help of generating functions based upon (1).

To simplify the notation somewhat I will often omit the argument and write e.g. $(\ln\circ g)^k$ instead of $\left(\ln(g(x))\right)^k$. Now, putting the Complete Bell polynomial in OPs question on the left hand side and the other terms to the right hand side we want to show

Following identity is valid:

\begin{align*} \sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln\circ g)}&=\sum_{k=1}^{n}(\ln\circ g)^{k}B_{n,k}^f\tag{5}\\ &\qquad+\sum_{k=1}^{n}\sum_{m=0}^{n-k}\sum_{j=0}^{m}\binom{m}{j} \frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{{n,m+k}_{(f\rightarrow g)^k}}^f\qquad\qquad n\geq 1 \end{align*}

Please note, the following abbreviations are used in (5) and the expressions below: \begin{align*} &f:= f(x), \qquad g := g(x), \qquad f_k := \frac{d^k}{dx^k}f(x), \qquad g_k :=\frac{d^k}{dx^k}g(x)\\ &d_k := \frac{d^k}{dx^k}\left(f(x)\ln(g(x)\right),\qquad\frac{d^j}{d(f)^j}:= \frac{d^j}{d(f(x))^j}\\ &(f)_k := f(x)\left(f(x)-1\right)\cdot\ldots\cdot\left(f(x)-k+1\right) \end{align*}

Using the generating function (1) we observe:

\begin{align*} \sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln\circ g)}&=n!\sum_{k=1}^n\frac{1}{k!}[t^n]\left(\sum_{m\geq 1}d_m\frac{t^m}{m!}\right)^k\\ \\ \sum_{k=1}^{n}(\ln\circ g)^{k}B_{n,k}^f&=n!\sum_{k=1}^{n}(\ln\circ g)^{k}\frac{1}{k!}[t^n]\left(\sum_{m\geq 1}f_m\frac{t^m}{m!}\right)^k \\ \\ \sum_{k=1}^{n}\sum_{m=0}^{n-k}\sum_{j=0}^{m}\binom{m}{j}& \frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{{n,m+k}_{(f\rightarrow g)^k}}^f\\ &=n!\sum_{k=1}^n\frac{1}{k!}\frac{1}{g^k}\sum_{m=0}^{n-k}\frac{1}{m!}\sum_{j=0}^m \binom{m}{j}\left(\ln\circ g\right)^{m-j}\frac{d^j}{d(f)^j}[(f)_k]\\ &\qquad\cdot[t^n]\left(\sum_{j\geq 1}f_{j}\frac{t^{j}}{j!}\right)\left(\sum_{j\geq 1}g_{j}\frac{t^{j}}{j!}\right) \end{align*}

Putting all together gives following reformulation of the identity:

\begin{align*} \sum_{k=1}^n&\frac{1}{k!}[t^n]\left(\sum_{m\geq 1}d_m\frac{t^m}{m!}\right)^k\\ &=\sum_{k=1}^{n}\left(\ln \circ g\right)^k\frac{1}{k!}[t^n]\left(\sum_{m\geq 1}f_m\frac{t^m}{m!}\right)^k\\ &\qquad+\sum_{k=1}^n\frac{1}{k!}\frac{1}{g^k}\sum_{m=0}^{n-k}\frac{1}{m!} \sum_{j=0}^m\binom{m}{j}\left(\ln \circ g\right)^{m-j}\frac{d^j}{d(f)^j}[(f)_k]\\ &\qquad\qquad\cdot[t^n]\left(\sum_{j\geq 1}f_{j}\frac{t^{j}}{j!}\right)\left(\sum_{j\geq 1}g_{j}\frac{t^{j}}{j!}\right) \end{align*}

Note: Maybe this alternative representation could help to show OPs identity.

Verification of the identity for $n=2,3$

In order to verify the identity for small $n$, we need some polynomials $B_{n,k}$ in variables $f_j$ and $g_j$ ($j$-th derivative of $f$ and $j$). We do so by applying the $\diamond$ operator to $a=(f_1,f_2,\ldots)$ and $b=(g_1,g_2,\ldots)$.

\begin{array}{rlllll} a^{1\diamond}&=\left(f_1,\right.&f_2,&f_3,&f_4,&\left.\ldots\right)\\ b^{1\diamond}&=\left(g_1,\right.&g_2,&g_3,&g_4,&\left.\ldots\right)\\ \\ a^{2\diamond}&=\left(0,\right.&2f_1^2,&6f_1f_2,&8f_1f_3+6f_2^2,&\left.\ldots\right)\\ a^{1\diamond}\diamond b^{1\diamond}&=\left(0,\right.&2f_1g_1,&3f_1g_2+3f_2g_1,& 4f_1g_3+6f_2g_2+4f_3g_1,&\left.\ldots\right)\\ b^{2\diamond}&=\left(0,\right.&2g_1^2,&6g_1g_2,&8g_1g_3+6g_2^2,&\left.\ldots\right)\\ \\ a^{3\diamond}&=\left(0,\right.&0,&6f_1^3,&36f_1^2f_2,&\left.\ldots\right)\\ a^{2\diamond}\diamond b^{1\diamond}&=\left(0,\right.&0,&6f_1^2g_1,&12f_1^2g_2+24f_1f_2g_1,&\left.\ldots\right)\\ a^{1\diamond}\diamond b^{2\diamond}&=\left(0,\right.&0,&6f_1g_1^2,&24f_1g_1g_2+12f_2g_1^2,&\left.\ldots\right)\\ b^{3\diamond}&=\left(0,\right.&0,&6g_1^3,&36g_1^2g_2,&\left.\ldots\right)\\ \end{array}

Case $n=2$:

Each of the three sums of the identity is calculated separately.

\begin{align*} \sum_{k=1}^{2}&B_{2,k}^{f\cdot(\ln \circ g)}\\ &=B_{2,1}^{f\cdot(\ln \circ g)}+B_{2,2}^{f\cdot(\ln \circ g)}\\ &=\frac{d^2}{{dx}^2}\left(f (\ln \circ g)\right)+\left(\frac{d}{dx}\left(f(\ln \circ g)\right)\right)^2\\ &=\left(f_2(\ln \circ g)+2f_1\frac{g_1}{g}+f\frac{g_2}{g}-f\frac{g_1^2}{g^2}\right)+\left(f_1(\ln \circ g)+f\frac{g_1}{g}\right)^2\\ &=\left(f_2(\ln \circ g)+2f_1\frac{g_1}{g}+f\frac{g_2}{g}-f\frac{g_1^2}{g^2}\right)+\left(f_1^2(\ln \circ g)^2+2ff_1(\ln \circ g)\frac{g_1}{g}+f^2\frac{g_1^2}{g^2}\right)\\ \\ \sum_{k=1}^{2}&(\ln \circ g)^kB_{2,k}^f\\ &=(\ln \circ g)B_{2,1}^f+(\ln \circ g)^2B_{2,2}^f\\ &=(\ln \circ g)f_2+(\ln \circ g)^2f_1^2\\ \\ \sum_{k=1}^{2}&\sum_{m=0}^{2-k}\sum_{j=0}^m\binom{m}{j} \frac{(\ln \circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[\left(f\right)_k]\frac{\left(a^{m_\diamond}\diamond b^{k_{\diamond}}\right)_2}{m!k!}\\ &=\sum_{k=1}^{2}\frac{1}{g^k}\sum_{m=0}^{2-k}\frac{\left(a^{m_\diamond}\diamond b^{k_{\diamond}}\right)_2}{m!k!} \sum_{j=0}^m\binom{m}{j}\frac{(\ln \circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[\left(f\right)_k]\\ &=\frac{1}{g}\sum_{m=0}^1\frac{\left(a^{m_\diamond}\diamond b^{1_{\diamond}}\right)_2}{m!1!} \sum_{j=0}^m\binom{m}{j}(\ln \circ g)^{m-j}\frac{d^j}{d(f)^j}(f)\\ &\qquad+\frac{1}{g^2}\sum_{m=0}^0\frac{\left(a^{m_\diamond}\diamond b^{2_{\diamond}}\right)_2}{m!2!} \sum_{j=0}^m\binom{m}{j}(\ln \circ g)^{m-j}\frac{d^j}{d(f)^j}\left(f(f-1)\right)\\ &=\frac{1}{g}\left[\frac{\left(b^{1_\diamond}\right)_2}{1!}\binom{0}{0}f+\frac{\left(a^{1_\diamond}\diamond b^{1_\diamond}\right)_2}{1!1!}\left(\binom{1}{0}(\ln \circ g)f+\binom{1}{1}\frac{d}{d(f)}f\right)\right]\\ &\qquad+\frac{1}{g^2}\left[\frac{\left(b^{2_\diamond}\right)_2}{2!}\binom{0}{0}f\left(f-1\right)\right]\\ &=\frac{1}{g}\left[g_2f+2f_1g_1\left((\ln \circ g) f+1\right)\right]+\frac{1}{g^2}\left[g_1^2f\left(f-1\right)\right] \end{align*}

Comparison of the results of these sums shows the validity of the claim: \begin{align*} \sum_{k=1}^{2}B_{2,k}^{f\cdot(\ln \circ g)}&=\sum_{k=1}^{2}(\ln \circ g)^kB_{2,k}^f\\ &\qquad+\sum_{k=1}^{2}\sum_{m=0}^{2-k}\sum_{j=0}^m\binom{m}{j} \frac{(\ln \circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[\left(f\right)_k]\frac{\left(a^{m_\diamond}\diamond b^{k_{\diamond}}\right)_2}{m!k!}\\ \end{align*}

Case $n=3$:

Each of the three sums of the identity is calculated separately.

\begin{align*} \sum_{k=1}^{3}&B_{3,k}^{f\cdot(\ln \circ g)}\\ &=B_{3,1}^{f\cdot(\ln \circ g)}+B_{3,2}^{f\cdot(\ln \circ g)}+B_{3,3}^{f\cdot(\ln \circ g)}\\ &=\frac{d^3}{{dx}^3}\left(f(\ln \circ g)\right)+3\frac{d}{dx}\left(f(\ln \circ g)\right)\frac{d^2}{{dx}^2}\left(f(\ln \circ g)\right) +\left(\frac{d}{dx}(\ln \circ g)\right)^3\\ &=\left(f_3(\ln \circ g)+3f_2\frac{g_1}{g}+3f_1\frac{g_2}{g}-3f_1\frac{g_1^2}{g^2}+f\frac{g_3}{g} +2f\frac{g_1^3}{g^3}-3f\frac{g_1g_2}{g^2}\right)\\ &\qquad+3\left(f_1(\ln \circ g)+f\frac{g_1}{g}\right)\left(f_2(\ln \circ g)+2f_1\frac{g_1}{g}+f\frac{g_2}{g}-f\frac{g_1^2}{g^2}\right)\\ &\qquad+\left(f_1(\ln \circ g)+f\frac{g_1}{g}\right)^3\\ &=\left(f_3(\ln \circ g)+3f_2\frac{g_1}{g}+3f_1\frac{g_2}{g}-3f_1\frac{g_1^2}{g^2}+f\frac{g_3}{g} +2f\frac{g_1^3}{g^3}-3f\frac{g_1g_2}{g^2}\right)\\ &\qquad+\left(3f_1f_2(\ln \circ g)^2+3ff_2(\ln \circ g)\frac{g_1}{g}+6f_1^2(\ln \circ g)\frac{g_1}{g}+6ff_1\frac{g_1^2}{g^2}\right.\\ &\qquad\qquad\left.+3ff_1(\ln \circ g)\frac{g_2}{g}+3f^2\frac{g_1g_2}{g^2}-3ff_1(\ln \circ g)\frac{g_1^2}{g^2}-3f^2\frac{g_1^3}{g^3}\right)\\ &\qquad+\left(f_1^3(\ln \circ g)^3+3ff_1^2(\ln \circ g)^2\frac{g_1}{g}+3f^2f_1(\ln \circ g)\frac{g_1^2}{g^2}+f^3\frac{g_1^3}{g^3}\right) \\ \\ \sum_{k=1}^{3}&(\ln \circ g)^kB_{3,k}^f\\ &=(\ln \circ g)B_{3,1}^f+(\ln \circ g)^2B_{3,2}^f+(\ln \circ g)^3B_{3,3}^f\\ &=(\ln \circ g)f_3+3(\ln \circ g)^2f_1f_2+(\ln \circ g)^3f_1^3 \end{align*} \begin{align*} \sum_{k=1}^{3}&\sum_{m=0}^{3-k}\sum_{j=0}^m\binom{m}{j} \frac{(\ln \circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[\left(f\right)_k]\frac{\left(a^{m_\diamond}\diamond b^{k_{\diamond}}\right)_3}{m!k!}\\ &=\sum_{k=1}^{3}\frac{1}{g^k}\sum_{m=0}^{3-k}\frac{\left(a^{m_\diamond}\diamond b^{k_{\diamond}}\right)_3}{m!k!} \sum_{j=0}^m\binom{m}{j}\frac{(\ln \circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[\left(f\right)_k]\\ &=\frac{1}{g}\sum_{m=0}^2\frac{\left(a^{m_\diamond}\diamond b^{1_{\diamond}}\right)_3}{m!1!} \sum_{j=0}^m\binom{m}{j}(\ln \circ g)^{m-j}\frac{d^j}{d(f)^j}(f)\\ &\qquad+\frac{1}{g^2}\sum_{m=0}^1\frac{\left(a^{m_\diamond}\diamond b^{2_{\diamond}}\right)_3}{m!2!} \sum_{j=0}^m\binom{m}{j}(\ln \circ g)^{m-j}\frac{d^j}{d(f)^j}\left(f(f-1)\right)\\ &\qquad+\frac{1}{g^3}\sum_{m=0}^0\frac{\left(a^{m_\diamond}\diamond b^{3_{\diamond}}\right)_3}{m!3!} \sum_{j=0}^m\binom{m}{j}(\ln \circ g)^{m-j}\frac{d^j}{d(f)^j}\left(f(f-1)(f-2)\right)\\ &=\frac{1}{g}\left[\frac{\left(b^{1_\diamond}\right)_3}{1!}\binom{0}{0}f +\frac{\left(a^{1_\diamond}\diamond b^{1_\diamond}\right)_3}{1!1!}\left[\binom{1}{0}(\ln \circ g) f +\binom{1}{1}\frac{d}{d(f)}f\right]\right.\\ &\qquad\qquad+\left.\frac{\left(a^{2_\diamond}\diamond b^{1_\diamond}\right)_3}{2!1!} \left[\binom{2}{0}(\ln \circ g)^2f+\binom{2}{1}(\ln \circ g)\frac{d}{df}f +\binom{2}{2}\frac{d^2}{{df}^2}f\right]\right]\\ &\qquad+\frac{1}{g^2}\left[\frac{\left(b^{2_\diamond}\right)_3}{2!}\binom{0}{0}f\left(f-1\right)\right.\\ &\qquad\qquad+\left.\frac{\left(a^{1_\diamond}\diamond b^{2_\diamond}\right)_3}{1!2!}\left[\binom{1}{0}(\ln \circ g) f(f-1) +\binom{1}{1}\frac{d}{d(f)}f(f-1)\right]\right]\\ &\qquad+\frac{1}{g^3}\left[\frac{\left(b^{3_\diamond}\right)_3}{3!}\binom{0}{0}f\left(f-1\right)(f-2)\right]\\ &=\frac{1}{g}\left[g_3f+\left(3f_1g_2+3f_2g_1\right)\left[(\ln \circ g)f+1\right] +\left(3f_1^2g_1\right)\left[(\ln \circ g)^2f+2(\ln \circ g)\right]\right]\\ &\qquad+\frac{1}{g^2}\left[3g_1g_2f\left(f-1\right) +3f_1g_1^2\left[(\ln \circ g) f(f-1)+2f-1\right]\right]\\ &\qquad+\frac{1}{g^3}\left[g_1^3f(f-1)(f-2)\right]\\ \end{align*}

Comparison of the results of these sums shows the validity of the claim: \begin{align*} \sum_{k=1}^{3}B_{3,k}^{f\cdot(\ln \circ g)}&=\sum_{k=1}^{3}(\ln \circ g)^kB_{3,k}^f\\ &\qquad+\sum_{k=1}^{3}\sum_{m=0}^{3-k}\sum_{j=0}^m\binom{m}{j} \frac{(\ln \circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[\left(f\right)_k]\frac{\left(a^{m_\diamond}\diamond b^{k_{\diamond}}\right)_3}{m!k!} \end{align*}

Note: This answer is aimed to provide a complete solution to this challenging problem. It is based upon a series of papers about compositae of functions by Vladimir Kruchinin and a paper about Bell polynomials by Warren P. Johnsen (see references at the end).

If this somewhat extensive elaboration is too cumbersome for a detailed reading, I suggest to read the overview and take instead a look at Manipulation of Bell Polynomials. This was my starter in applying the techniques used here. Another way to go through this answer is to focus on highlighted regions and skip this way less important details.

Note: [2015-03-08]

It's a great pleasure to provide now a complete answer to OPs challenging question. I could close the gap by adding Step 2 and Step 3 of Part 3 of my answer. It was a really tough job to perform all the calculations (and much more verification steps behind the scene). So, if the interested reader is willing to review parts of this answer any fruitful hints to reduce the size of this proof are appreciated.

Overview:

The answer is divided into three parts.

Part 1: Convenient representation of OPs formula

We transform OPs expression conveniently and replace the convolution product $a\diamond b$ with corresponding partial Bell polynomials to ease further calculations. We show

OPs formula is equivalent to (argument $x$ omitted):

\begin{align*} \sum_{k=1}^{n}&B_{n,k}^{f\cdot (\ln\circ g)}= \sum_{k=1}^{n}\left((\ln\circ g)^{k}B_{n,k}^f+\frac{(f)_k}{g^k}B_{n,k}^g\right)\tag{1}\\ &+\sum_{l=1}^{n-1}\sum_{m=1}^{l}\sum_{k=1}^{n-l}\sum_{j=0}^{m}\binom{n}{l}\binom{m}{j} \frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{l,m}^{f}B_{n-l,k}^{g}\qquad\qquad n\geq 1 \end{align*}

Part 2: Derivation of $B_{n,k}^{f\cdot (\ln\circ g)}$

The strategy to proof OPs formula (1) is to derive the partial Bell polynomials for $B_{n,k}^{f\cdot (\ln\circ g)}$ from scratch and show that the so resulting formula is the same as (1). We do so by starting with partial Bell polynomials of $f$ and $g$ and building stepwise more complex partial Bell polynomials.

2.1 Preliminaries:

We will recognise that deriving Bell polynomials is strongly related to composites of generating functions and their Taylor expansion series. We introduce the necessary concepts and notation.

2.2 From $B_{n,k}^{\ln}$ and $B_{n,k}^g$ to $B_{n,k}^{\ln \circ g}$

We create the partial Bell polynomials $B_{n,k}^{\ln \circ g}$ from $B_{n,k}^{\ln}$ and $B_{n,k}^g$ and show

The following is valid \begin{align*} B^{\ln\circ g}_{n,k}&=\sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}\qquad 1\leq k \leq n\tag{2} \end{align*}

with $\begin{bmatrix}j\\k\end{bmatrix}$ the unsigned Stirling numbers of the first kind.

2.3 From $B_{n,k}^{f}$ and $B_{n,k}^h$ to $B_{n,k}^{f\cdot h}$

We create the partial Bell polynomials $B_{n,k}^{f \cdot h}$ from $B_{n,k}^{f}$ and $B_{n,k}^h$ and show

The following is valid

\begin{align*} B_{n,k}^{f\cdot h}&=h^k B_{n,k}^{f} + f^k B_{n,k}^{h}\\ &\qquad+\frac{1}{k!}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}m_{1}! m_{2}!f^{k-m_1}h^{k-m_2}\tag{3}\\ &\qquad\qquad\cdot \sum_{l=1}^{n-1}\binom{n}{l}B_{l,m_1}^fB_{n-l,m_2}^{h}\qquad\qquad 1\leq k \leq n\\ \end{align*}

We thereby find some nice combinatorial identities and use interesting techniques (by Egorychev) in order to prove them.

2.4 From $B_{n,k}^{f \cdot h}$ and $B_{n,k}^{\ln \circ g}$ to $B_{n,k}^{f\cdot (\ln \circ g)}$

We combine the partial Bell polynomials $B_{n,k}^{f \cdot h}$ and $B_{n,k}^{\ln \circ g}$ and summing from $1 \leq k \leq n$ results in

\begin{align*} \sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln \circ g)} &=\sum_{k=1}^{n}\left((\ln \circ g)^k B_{n,k}^{f} + f^k \sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}\right)\\ &\qquad+\sum_{k=1}^{n}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\frac{m_2!}{(k-m_1)!}\binom{m_1}{k-m_2}f^{k-m_1} (\ln \circ g)^{k-m_2}\\ &\qquad\qquad\cdot\sum_{l=1}^{n-1}\sum_{j=m_2}^{n-l}\frac{(-1)^{j-m_2}}{g^j}\begin{bmatrix}j\\m_2\end{bmatrix}\binom{n}{l}B_{l,m_1}^fB^{g}_{n-l,j}\tag{4} \end{align*}

Part 3:

We have now derived an identity for $\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln \circ g)}$. We finally show that this identity (4) is equivalent to OPs formula (1) and the proof is completed.

Solution:

Part 1:

First we take a closer look at the structure of OPs expression. In order to do so we introduce the abbreviation:

$$B_{n,k}^{f}(x) := B_{n,k}(f^\prime(x),f^{\prime\prime}(x),\ldots,f^{(n-k+1)})$$

OPs expression can therefore be written as

\begin{align*} \sum_{k=1}^{n}&\ln^{k}(g(x))B_{n,k}^f(x)=\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln\circ g)}(x)\\ &-\sum_{k=1}^{n}\sum_{m=0}^{n-k}\sum_{j=0}^{m}\binom{m}{j} \frac{\ln^{m-j}(g(x))}{g(x)^k}\frac{d^j}{d(f(x))^j}[(f(x))_k]B_{{n,m+k}_{(f\rightarrow g)^k}}^f(x) \end{align*}

We see partial Bell polynomials with different complexity. The most complex is $B_{n,k}^{f\cdot (\ln \circ g)}$ which is based upon a composition of $\ln$ with a function $g$ and a multiplication with a function $f$. Next we have $B_{{n,m+k}_{(f\rightarrow g)^k}}^f$, which is an expression containing the Bell polynomials $B_{n,k}^f$ and $B_{n,k}^g$ and is therefore based upon $f$ and $g$. The simplest one is $B_{n,k}^{f}$ based upon $f$ only. Now we reorganise the expression and put thereby $B_{n,k}^{f\cdot (\ln \circ g)}$ on the LHS.

In order to keep the complicated expressions better manageable, we will also sometimes omit the variable $x$. OPs expression can now be written as:

\begin{align*} \sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln\circ g)}=\sum_{k=1}^{n}&(\ln\circ g)^{k}B_{n,k}^f\tag{5}\\ &+\sum_{k=1}^{n}\sum_{m=0}^{n-k}\sum_{j=0}^{m}\binom{m}{j} \frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{{n,m+k}_{(f\rightarrow g)^k}}^f\qquad\qquad n\geq 1 \end{align*}

The next step is to take a closer look at the convolution product \begin{align*} B_{{n,k}_{(f\rightarrow g)^c}}^f:= \frac{(a^{(k-c)\diamond} \diamond b^{c\diamond})_n}{(k-c)!c!}\qquad 1\leq k\leq n, 0\leq c\leq k \end{align*} with $a=(f^{\prime},f^{\prime\prime},\ldots)$ and $b=(g^{\prime},g^{\prime\prime},\ldots)$

Note: You might have a look at my first answer to this question. See the section: The convolution product $x \diamond y$ demystified which provides some information about it.

We note that the partial Bell polynomial $B_{n,k}^f$ is defined as:

\begin{align*} \frac{1}{n!}B_{n,k}^f=[t^n]\frac{1}{k!}\left(\sum_{m\geq 1}f^{(m)}\frac{t^m}{m!}\right)^k\qquad 1 \leq k \leq n \end{align*}

You might also have a look at the section: Definition of partial Bell polynomials $B_{n,k}$ of my first answer for more information.

With this information at hand we can show that

the following is valid for $1\leq k \leq n$:

\begin{equation*} B_{{n,k}_{(f\rightarrow g)^c}}^f= \begin{cases} B_{n,k}^f\qquad& c=0\\ \sum_{l=1}^{n-1}\binom{n}{l}B_{l,k-c}^fB_{n-l,c}^g\qquad& n>1, 0 < c < k\tag{6}\\ B_{n,k}^g\qquad& c=k\\ \end{cases} \end{equation*}

We observe for $n>1,0<c<k$

\begin{align*} \frac{1}{n!}B_{{n,k}_{(f\rightarrow g)^c}}^f&=[t^n]\frac{1}{(k-c)!}\left(\sum_{m\geq 1}f^{(m)}\frac{t^m}{m!}\right)^{k-c} \frac{1}{c!}\left(\sum_{m\geq 1}g^{(m)}\frac{t^m}{m!}\right)^{c}\\ &=\sum_{l=1}^{n-1}[t^l]\frac{1}{(k-c)!}\left(\sum_{m\geq 1}f^{(m)}\frac{t^m}{m!}\right)^{k-c} [t^{n-l}]\frac{1}{c!}\left(\sum_{m\geq 1}g^{(m)}\frac{t^m}{m!}\right)^{c}\\ &=\sum_{l=1}^{n-1}\frac{1}{l!}B_{l,k-c}^{f}\frac{1}{(n-l)!}B_{n-l,c}^{g}\\ &=\frac{1}{n!}\sum_{l=1}^{n-1}\binom{n}{l}B_{l,k-c}^{f}B_{n-l,c}^{g}\\ \end{align*}

The cases $n=1$ and $c\in \{0,k\}$ can be easily verified and (6) follows.

With the help of (6) we can now write OPs expression (5) as:

\begin{align*} \sum_{k=1}^{n}&B_{n,k}^{f\cdot (\ln\circ g)}= \sum_{k=1}^{n}\left((\ln\circ g)^{k}B_{n,k}^f+\frac{(f)_k}{g^k}B_{n,k}^g\right)\tag{7}\\ &+\sum_{k=1}^{n}\sum_{m=1}^{n-k}\sum_{j=0}^{m}\binom{m}{j} \frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]\sum_{l=1}^{n-1}\binom{n}{l}B_{l,m}^{f}B_{n-l,k}^{g}\qquad\qquad n\geq 1 \end{align*}

Observe that in (7) the index range of $m$ starts with $1$ instead of $0$. The case $m=0$ is now treated as $\sum_{k=1}^{n}\frac{(f)_k}{g^k}B_{n,k}^g$ corresponding to the case $c=k$ in (6) whereas the other summand $\sum_{k=1}^{n}(\ln\circ g)^{k}B_{n,k}^f$ corresponds to the case $c=0$.

We note that $B_{l,m}^{f}=0$ for $m>l$ and $B_{n-l,k}^{g}=0$ for $k>n-l$. In order to analyse the sum in (7) conveniently we now restrict the index range precisely to those values, which give a contribution to the sum.

Relevant for these considerations is the expression

\begin{align*} \sum_{k=1}^{n-1}&\sum_{l=1}^{n-1}\sum_{m=1}^{n-k}B_{l,m}^fB_{n-l,k}^g\tag{8}\\ &=\sum_{k=1}^{n-1}\sum_{l=1}^{n-k}\sum_{m=1}^{n-k}B_{l,m}^fB_{n-l,k}^g\tag{9}\\ &=\sum_{k=1}^{n-1}\sum_{l=1}^{n-k}\sum_{m=1}^{l}B_{l,m}^fB_{n-l,k}^g\tag{10}\\ &=\sum_{l=1}^{n-1}\sum_{k=1}^{n-l}\sum_{m=1}^{l}B_{l,m}^fB_{n-l,k}^g\tag{11}\\ \end{align*}

Comment:

• (8) Index $k\leq n-1$ since $(l\geq 1$ and $k=n)\Rightarrow (B_{n-l,k}^g=0)$

• (9) Index $l\leq n-k$ since otherwise $B_{n-l,k}^g=0$.

• (10) Index $m\leq l$ since otherwise $B_{l,m}^g=0$.

• (11) Exchange sums with index $l$ and $k$

OPs expression can therefore be written as:

\begin{align*} \sum_{k=1}^{n}&B_{n,k}^{f\cdot (\ln\circ g)}= \sum_{k=1}^{n}\left((\ln\circ g)^{k}B_{n,k}^f+\frac{(f)_k}{g^k}B_{n,k}^g\right)\tag{12}\\ &+\sum_{l=1}^{n-1}\sum_{m=1}^{l}\sum_{k=1}^{n-l}\sum_{j=0}^{m}\binom{n}{l}\binom{m}{j} \frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{l,m}^{f}B_{n-l,k}^{g}\qquad\qquad n\geq 1 \end{align*}

Summary of part 1: We have now finalised this part. OPs expression in the form (12) is a convenient representation for further analysis.

And now for something completely different: In the following part 2 we develop a formula for $$\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln\circ g)}$$ from scratch. In the last part 3 we will show that both formulae are equal and the job is done.

Part 2:

The development of a formula for $\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln\circ g)}$ from scratch is done in several steps. We start with some

Part 2.1: Preliminaries

Let's consider a function $f=f(z)$ and its Taylor series expansion at a point $x$ \begin{align*} f(z+x)=\sum_{n\geq 0}\frac{f^{(n)}(x)}{n!}z^n \end{align*} We will take a look at the $k$-th power of the difference $f(x+z)-f(x)$ since there is a strong relationship with the partial Bell Polynomials $B_{n,k}\ \ (1\leq k\leq n)$: \begin{align*} B_{n,k}(f^{\prime},f^{\prime\prime},\ldots,f^{(n-k+1)}) &=\frac{1}{k!}\sum_{\pi_k \in C_n}\binom{n}{\lambda_1,\lambda_2,\ldots,\lambda_k}f^{(\lambda_1)}f^{(\lambda_2)}\cdot\ldots\cdot f^{(\lambda_n)}\qquad\\ &=\frac{n!}{k!}\sum_{\pi_k \in C_n}\frac{f^{(\lambda_1)}(x)}{\lambda_1!}\frac{f^{(\lambda_2)}(x)}{\lambda_2!}\cdot\ldots\cdot\frac{f^{(\lambda_n)}(x)}{\lambda_n!} \end{align*} Here $f^{(n)}$ is the $n$-th derivative of $f=f(x)$, $C_n$ is the set of compositions of $n$ and $\pi_k$ is the composition of $n$ with $k$ parts exactly $\lambda_1+\lambda_2+\ldots+\lambda_k=n$.

In order to do so its convenient to define \begin{align*} Y_f(x,z) := f(x+z)-f(x) =\sum_{n> 0}\frac{f^{(n)}(x)}{n!}z^n \end{align*} and consider the $k$-th power of $Y_f(x,z)$. We get \begin{align*} \left(Y_f(x,z)\right)^k=\left(f(x+z)-f(x)\right)^k=\sum_{n\geq k}Y_f^{\triangle}(n,k,x)z^n \end{align*} $Y_f^{\triangle}(n,k,x)$ is introduced to denote the coefficient of $z^n$ of $\left(Y_f(x,z)\right)^k$. In the following we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a power series. We observe

\begin{align*} Y_f^\triangle(n,k,x)&=[z^n]\left(f(x+z)-f(x)\right)^k\\ &=\sum_{\pi_{k}\in C_n}\frac{f^{(\lambda_1)}(x)}{\lambda_1!}\frac{f^{(\lambda_2)}(x)}{\lambda_2!}\cdot\ldots\cdot\frac{f^{(\lambda_n)}(x)}{\lambda_n!}\\ &=\frac{k!}{n!}B_{n,k}(f^{\prime},f^{\prime\prime},\ldots,f^{(n-k+1)})\qquad\qquad 1\leq k \leq n \end{align*} It follows \begin{align*} B_{n,k}(f^{\prime},f^{\prime\prime},\ldots,f^{(n-k+1)})=\frac{n!}{k!}Y_f^\triangle(n,k,x) \end{align*}

We again use following abbreviation for the $B_{n,k}$: \begin{align*} B^{f}_{n,k}(x):=B_{n,k}(f^{\prime},f^{\prime\prime},\ldots,f^{(n-k+1)}) \end{align*}

Part 2.2: Calculation of $B^{\ln \circ g}_{n,k}$

Using the notation introduced in part 2.1 we consider the difference of the composita of $(\ln \circ g)$: $$Y_{\ln \circ g}(x,z)=\ln(g(x+z))-\ln(g(x))=\sum_{n>0}\frac{(\ln \circ g)^{(n)}(x)}{n!}z^n$$ and calculate the coefficient $Y^\triangle_{\ln \circ g}(n,k,x)$ of $z^n$ of the $k$-th powers of the composita $Y^{\ln \circ g}(x,z), k\geq 1$.

We do it in three steps. First we calculate $B^{g}_{n,k}$, then $B^{\ln}_{n,k}$ and finally we get $B^{\ln \circ g}_{n,k}$.

Step 2.2.1: Calculation of $B^{g}_{n,k}$

\begin{align*} Y_g(x,z)&=g(x+z)-g(x)=\sum_{n>0}\frac{g^{(n)}(x)}{n!}z^n\\ \left(Y_g(x,z)\right)^k&=\left(g(x+z)-g(x)\right)^k=\sum_{n\geq k}Y_g^{\triangle}(n,k,x)z^n\\ \\ Y_g^{\triangle}(n,k,x)&=[z^n]\left(g(x+z)-g(x)\right)^k\\ &=\frac{k!}{n!}B^g_{n,k}(x) \end{align*}

It follows \begin{align*} B^g_{n,k}(x)=\frac{n!}{k!}Y_g^{\triangle}(n,k,x) \end{align*}

Step 2.2.2: Calculation of $B^{\ln}_{n,k}$

\begin{align*} Y_{\ln}&=\ln(x+z)-\ln(x)=\sum_{n>0}\frac{\ln^{(n)}(x)}{n!}z^n\\ \left(Y_{\ln}\right)^k&=\left(\ln(x+z)-\ln(x)\right)^k=\sum_{n\geq k}Y_{\ln}^{\triangle}(n,k,x)z^n\\ \\ Y_{\ln}^{\triangle}(n,k,x)&=[z^n]\left(\ln(x+z)-\ln(x)\right)^k\\ &=[z^n]\left(\ln\left(1+\frac{z}{x}\right)\right)^k\\ &=[z^n]\sum_{n\geq k}k!(-1)^{n-k} \begin{bmatrix}n\\k\end{bmatrix} x^{-n}\frac{z^n}{n!}\tag{13}\\ &=\frac{k!}{n!}(-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix}x^{-n}\\ &=\frac{k!}{n!}B^{\ln}_{n,k}(x) \end{align*}

It follows \begin{align*} B^{\ln}_{n,k}(x)= (-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix} x^{-n} \end{align*} In (13) we use the unsigned Stirling numbers of the first kind $\begin{bmatrix}n\\k\end{bmatrix}$, which are defined as coefficients in the exponential power series for the $k$-th powers of the logarithm $\ln(1+z)$: $$\sum_{n\geq k}(-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix}\frac{z^n}{n!}=\frac{1}{k!}\left(\ln (1+z)\right)^k$$

Step 2.2.3: Calculation of $B^{\ln\circ g}_{n,k}$ \begin{align*} Y_{\ln\circ g}(x,z)&=\ln\left(g(x+z)\right)-\ln\left(g(x)\right)=\sum_{n>0}\frac{(\ln\circ g)^{(n)}(x)}{n!}z^n\\ \left(Y_{\ln\circ g}(x,z)\right)^k&=\left(\ln\left(g(x+z)\right)-\ln\left(g(x)\right)\right)^k=\sum_{n\geq k}Y_{\ln\circ g}^{\triangle}(n,k,x)z^n\\ \end{align*}

According to Theorem 6 in Kruch_3 the composita of $(\ln \circ g)$ are given as \begin{align*} Y_{\ln\circ g}^{\triangle}(n,k,x)=\sum_{j=k}^{n}Y_{g}^{\triangle}(n,j,x)\cdot Y_{\ln}^{\triangle}(j,k,g(x)) \end{align*} It follows \begin{align*} \frac{k!}{n!}B^{\ln\circ g}_{n,k}(x)&=\sum_{j=k}^{n}\frac{j!}{n!}B^{g}_{n,j}(x) \cdot\frac{k!}{j!}B^{\ln}_{j,k}(g(x))\\ &=\sum_{j=k}^{n}\frac{k!}{n!}B^{g}_{n,j}(x)(-1)^{j-k}\begin{bmatrix}j\\k\end{bmatrix}\left(g(x)\right)^{-j} \end{align*} We therefore get \begin{align*} B^{\ln\circ g}_{n,k}(x)&=\sum_{j=k}^{n}B^{g}_{n,j}(x)(-1)^{j-k}\begin{bmatrix}j\\k\end{bmatrix}\left(g(x)\right)^{-j}\tag{14} \end{align*}

The next step is to calculate $B_{n,k}$ for the multiplication of $f\cdot h$ of two functions $f$ and $h$.

Part 2.3: Calculation of $B^{f\cdot h}_{n,k}$

We start similarly as in part 2.2:

Step 2.3.1: Derivation of $B^{f\cdot h}_{n,k}$

\begin{align*} Y_{f\cdot h}(x,z)&=f(x+z)h(x+z)-f(x)h(x)=\sum_{n>0}\frac{(f\cdot h)^{(n)}(x)}{n!}z^n\\ \left(Y_{f\cdot h}(x,z)\right)^k&=\left(f(x+z)h(x+z)-f(x)h(x)\right)^k=\sum_{n\geq k}Y_{f\cdot h}^{\triangle}(n,k,x)z^n\\ \end{align*}

\begin{align*} Y_{f\cdot h}^{\triangle}(n,k,x)&=[z^n]\left(f(x+z)h(x+z)-f(x)h(x)\right)^k\\ &=[z^n]\sum_{j=0}^{k}\binom{k}{j}\left(f(x+z)h(x+z)\right)^j\left(-f(x)h(x)\right)^{k-j}\\ &=\sum_{j=0}^{k}\binom{k}{j}\left(-f(x)h(x)\right)^{k-j}[z^n]\left(f(x+z)h(x+z)\right)^j\\ &=\sum_{j=0}^{k}\binom{k}{j}\left(-f(x)h(x)\right)^{k-j}\\ &\qquad\cdot\sum_{i=0}^{n}\left([z^i]\left(f(x+z)\right)^j\right)\left([z^{n-i}]\left(h(x+z)\right)^j\right)\tag{15} \end{align*} We want to replace the expressions $[z^n]\left(f(x+z)\right)^j$ in (15) with an expression containing $Y_{f}(x,z)$ instead. Then we can again substitute $Y_f(x,z)$ with the corresponding $B_{n,k}^f$. We observe: \begin{align*} [z^n]f(x+z)^k&=[z^n]\left(f(x+z)-f(x)+f(x)\right)^k\\ &=[z^n]\sum_{j=0}^{k}\binom{k}{j}\left(f(x+z)-f(x)\right)^j\left(f(x)\right)^{k-j}\\ &=\left(f(x)\right)^k\delta_{n,0}+\sum_{j=1}^{k}\binom{k}{j}\left(f(x)\right)^{k-j}[z^n]\left(f(x+z)-f(x)\right)^j\\ &=\left(f(x)\right)^k\delta_{n,0}+\sum_{j=1}^{k}\binom{k}{j}\left(f(x)\right)^{k-j}Y_f^\triangle(n,j,x)\tag{16} \end{align*} Since $Y_f^\triangle(n,j,x)$ is defined for $n\geq 1$ only we have to consider the case $j=0$ separately. We use thereby the Kronecker $\delta$-symbol. Now substituting (16) in (15) gives:

\begin{align*} Y_{f\cdot h}^{\triangle}(n,k,x)&=\sum_{j=0}^{k}\binom{k}{j}\left(-f(x)h(x)\right)^{k-j}\sum_{l=0}^{n}\\ &\qquad\left(\left(f(x)\right)^j\delta_{l,0}+\sum_{m_1=0}^{j}\binom{j}{m_1}\left(f(x)\right)^{j-m_1}Y_f^\triangle(l,m_1,x)\right)\\ &\qquad\cdot\left(\left(h(x)\right)^j\delta_{n-l,0}+\sum_{m_2=0}^{j}\binom{j}{m_2}\left(h(x)\right)^{j-m_2}Y_h^\triangle(n-l,m_2,x)\right)\\ &=\left(f(x)\right)^k\sum_{j=0}^{k}\binom{k}{j}\left(-1\right)^{k-j}\sum_{m_2=1}^{j}\binom{j}{m_2}\left(h(x)\right)^{k-m_2}Y_h^\triangle(n,m_2,x)\\ &\qquad+\sum_{j=0}^{k}\binom{k}{j}(-1)^{k-j}\sum_{l=1}^{n-1} \left(\sum_{m_1=0}^{j}\binom{j}{m_1}\left(f(x)\right)^{k-m_1}Y_f^\triangle(l,m_1,x)\right)\\ &\qquad\qquad\cdot\left(\sum_{m_2=0}^{j}\binom{j}{m_2}\left(h(x)\right)^{k-m_2}Y_h^\triangle(n-l,m_2,x)\right)\\ &\qquad+\left(h(x)\right)^k\sum_{j=0}^{k}\binom{k}{j}\left(-1\right)^{k-j}\sum_{m_1=1}^{j}\binom{j}{m_1}\left(f(x)\right)^{k-m_1}Y_f^\triangle(n,m_1,x)\\ \end{align*}

Substituting $B_{n,k}^{f\cdot h}(x)$ for $Y_{f\cdot h}^{\triangle}(n,k,x)$ (and omitting the argument $x$) gives:

\begin{align*} k!\frac{B_{n,k}^{f\cdot h}}{(fh)^{k}}&=\sum_{j=1}^{k}\binom{k}{j}(-1)^{k-j} \left(\sum_{m=1}^j\binom{j}{m}m!\left(\frac{B_{n,m}^{h}}{h^{m}} + \frac{B_{n,m}^{f}}{f^{m}}\right)\right.\\ &\qquad+\sum_{l=1}^{n-1}\binom{n}{l}\left(\sum_{m_1=1}^j\binom{j}{m_1}m_{1}!\frac{B_{l,m_1}^f}{f^{m_1}}\right) \left.\left(\sum_{m_2=1}^j\binom{j}{m_2}m_{2}!\frac{B_{n-l,m_2}^h}{h^{m_2}}\right)\right)\tag{17}\\ \end{align*} with the factor $\frac{k!}{(fh)^k}$ on the left-hand side to better show the nice symmetry.

Step 2.3.2: Simplification of $B^{f\cdot h}_{n,k}$

We can simplify (17) considerably. Observe, that the RHS consists of two parts, the first being a double sum with structure

\begin{align*} \sum_{j=1}^{k}\binom{k}{j}(-1)^{k-j}\sum_{m=1}^j\binom{j}{m}\varphi_m \qquad\qquad k\geq 1 \end{align*} and $\varphi_m$ equal to \begin{align*} \varphi_m=m!\left(\frac{B_{n,m}^{f}}{f^{m}} + \frac{B_{n,m}^{h}}{h^{m}}\right)\tag{18} \end{align*} We show that following is valid

\begin{align*} \sum_{j=1}^{k}\binom{k}{j}(-1)^{k-j}\sum_{m=1}^j\binom{j}{m}\varphi_m=\varphi_k \qquad\qquad k\geq 1\tag{19} \end{align*}

We observe \begin{align*} \sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m=1}^j\binom{j}{m}\varphi_m\\ &=\sum_{m=1}^k\varphi_m\sum_{j=m}^{k}\binom{k}{j}\binom{j}{m}(-1)^{k-j}\\ &=\sum_{m=1}^k\varphi_m\sum_{j=m}^{k}\binom{k}{m}\binom{k-m}{j-m}(-1)^{k-j}\\ &=\sum_{m=1}^k\binom{k}{m}\varphi_m\sum_{j=0}^{k-m}\binom{k-m}{j}(-1)^{(k-m)-j}\\ &=\sum_{m=1}^k\binom{k}{m}\varphi_m\delta_{{k-m},0}\\ &=\varphi_k \end{align*} and (19) follows.

Now we take a look at the second part of the RHS of (17). We identify following structure \begin{align*} \sum_{j=1}^{k}\binom{k}{j}(-1)^{k-j}\sum_{m_1=1}^j\binom{j}{m_1}\psi_{m_1}\sum_{m_2=1}^j\binom{j}{m_2}\psi_{m_2} \qquad\qquad k\geq 1 \end{align*}

with $\psi_{m_1},\psi_{m_2}$ equal to \begin{align*} \psi_{m_1}=m_{1}!\frac{B_{i,m_1}^f}{f^{m_1}}\qquad\qquad\psi_{m_2}=m_{2}!\frac{B_{n-i,m_2}^{h}}{h^{m_2}}\tag{20} \end{align*} The sum $\sum_{i=1}^{n-1}\binom{n}{i}$ is not part of our consideration. It does neither depend on $\psi_{m_1}$ nor on $\psi_{m_2}$ and we think of it as being at the front of the RHS (outside from $\sum_{j=1}^{k}$).

We show the following is valid

\begin{align*} \sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m_1=1}^j\binom{j}{m_1}\psi_{m_1}\sum_{m_2=1}^j\binom{j}{m_2}\psi_{m_2}\\ &\quad=\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}\psi_{m_1}\psi_{m_2}\qquad\qquad k\geq 1\tag{21} \end{align*} This identity is less simple than the identity (19) above but with the proper tool at hand we can prove this identity en passant. We will use Egorychevs residual calculus of formal power series.

Note: This powerful technique is based upon Cauchys residue theorem and was introduced by G.P. Egorychev (Integral Representation and the Computation of Combinatorial Sums) to compute binomial identies.

We use only two aspects of this theory:

Let $A(z)=\sum_{j=0}^{\infty}a_jz^j$ be a formal power series, then

• Write the binomial coeffients as residuals of corresponding formal power series

\begin{align*} \mathop{res}_{z}\frac{A(z)}{z^{j+1}}=a_j\tag{22} \end{align*}

• Apply the substitution rule for formal power series:

\begin{align*} A(z)=\sum_{j=0}^{\infty}a_jz^{j}=\sum_{j=0}^{\infty}z^j\mathop{res}_{w}\frac{A(w)}{w^{j+1}}\tag{23} \end{align*}

We start the proof with elementary transformations:

\begin{align*} \sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m_1=1}^j\binom{j}{m_1}\psi_{m_1}\sum_{m_2=1}^j\binom{j}{m_2}\psi_{m_2}\\ &=\sum_{m_1=1}^{k}\psi_{m_1}\sum_{j=m_1}^k\binom{k}{j}\binom{j}{m_1}(-1)^{k-j}\sum_{m_2=1}^j\binom{j}{m_2}\psi_{m_2}\tag{24}\\ &=\sum_{m_1=1}^{k}\psi_{m_1}\left(\sum_{m_2=1}^{m_1}\psi_{m_2}\sum_{j=m_1}^k\binom{k}{j}\binom{j}{m_1}\binom{j}{m_2}(-1)^{k-j}\right.\\ &\qquad\qquad\left.+\sum_{m_2=m_1+1}^{k}\psi_{m_2}\sum_{j=m_2}^k\binom{k}{j}\binom{j}{m_1}\binom{j}{m_2}(-1)^{k-j}\right)\tag{25}\\ &=\sum_{m_1=1}^{k}\psi_{m_1}\left(\sum_{m_2=1}^{m_1}\psi_{m_2}\sum_{j=m_1}^k\binom{k}{m_1}\binom{k-m_1}{j-m_1}\binom{j}{m_2}(-1)^{k-j}\right.\\ &\qquad\qquad\left.+\sum_{m_2=m_1+1}^{k}\psi_{m_2}\sum_{j=m_2}^k\binom{k}{m_2}\binom{k-m_2}{j-m_2}\binom{j}{m_1}(-1)^{k-j}\right)\tag{26}\\ &=\sum_{m_1=1}^{k}\psi_{m_1}\left(\binom{k}{m_1}\sum_{m_2=1}^{m_1}\psi_{m_2}\sum_{j=0}^{k-m_1}\binom{k-m_1}{j}\binom{j+m_1}{m_2}(-1)^{(k-m_1)-j}\right.\\ &\qquad\qquad\left.+\sum_{m_2=m_1+1}^{k}\binom{k}{m_2}\psi_{m_2}\sum_{j=0}^{k-m_2}\binom{k-m_2}{j}\binom{j+m_2}{m_1}(-1)^{(k-m_2)-j}\right)\tag{27}\\ \end{align*}

Comment:

• (24) Exchange of the sums with indices $j$ and $m_1$.

• (25) Exchange of the sums with indices $j$ and $m_2$.

• (26) Usage of the binomial identity $\binom{k}{j}\binom{j}{m}=\binom{k}{m}\binom{k-m}{j-m}$

• (27) Index shift $j\rightarrow j-m_1,j\rightarrow j-m_2$

We now apply Egorychevs technique to show that

The following is valid (and symmetrically with $m_1,m_2$ exchanged): \begin{align*} \sum_{j=0}^{k-m_1}\binom{k-m_1}{j}\binom{j+m_1}{m_2}(-1)^{(k-m_1)-j}=\binom{m_1}{k-m_2}\qquad\qquad k\geq 1\tag{28} \end{align*} We observe \begin{align*} \sum_{j=0}^{k-m_1}&\binom{k-m_1}{j}\binom{j+m_1}{m_2}(-1)^{(k-m_1)-j}\\ &=\sum_{j\geq 0}\binom{k-m_1}{j}\binom{-(m_2+1)}{j+m_1-m_2}(-1)^{k-m_2}\tag{29}\\ &=(-1)^{k-m_2}\sum_{j\geq 0}\mathop{res}_{z}\frac{(1+z)^{k-m_1}}{z^{j+1}} \mathop{res}_u\frac{(1+u)^{-(m_2+1)}}{u^{j+m_1-m_2+1}}\tag{30}\\ &=(-1)^{k-m_2}\mathop{res}_u\frac{(1+u)^{-(m_2+1)}}{u^{m_1-m_2+1}}\sum_{j\geq 0}u^{-j}\mathop{res}_{z}\frac{(1+z)^{k-m_1}}{z^{j+1}}\tag{31}\\ &=(-1)^{k-m_2}\mathop{res}_u\frac{(1+u)^{-(m_2+1)}}{u^{m_1-m_2+1}}\left(1+\frac{1}{u}\right)^{k-m_1}\tag{32}\\ &=(-1)^{k-m_2}\mathop{res}_u\frac{(1+u)^{k-m_2-m_1-1}}{u^{k-m_2+1}}\tag{33}\\ &=(-1)^{k-m_2}\binom{k-m_2-m_1-1}{k-m_2}\tag{34}\\ &=\binom{m_1}{k-m_2}\tag{35} \end{align*}

Comment:

• In (28) we use $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$ and changed the limit to $\infty$ without changing the value, since we add only $0$.

• In (29) we use $(1+w)^n$ as formal power series with the coefficients $\binom{n}{k}$ according to (22). Observe that $\binom{k-m_2}{j}$ is the coefficient of $[z^{j}]$ in $(1+z)^{k-m_1}$ and similarly for $u$.

• In (30) we do some rearrangement to prepare the application of the substitution rule

• In (31) we apply the substitution rule according to (23)

• In (32) there is a simple rearrangement

• In (33) we do the same as we did from (29) to (30) but in reverse direction.

We now proceed substituting the identity (28) in (27) and get:

\begin{align*} \sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m_1=1}^j\binom{j}{m_1}\psi_{m_1}\sum_{m_2=1}^j\binom{j}{m_2}\psi_{m_2}\\ &=\sum_{m_1=1}^{k}\left(\sum_{m_2=1}^{m_1}\binom{k}{m_1}\binom{m_1}{k-m_2}\psi_{m_1}\psi_{m_2}\right.\\ &\qquad\left.+\sum_{m_2=m_1+1}^k\binom{k}{m_2}\binom{m_2}{k-m_1}\psi_{m_1}\psi_{m_2}\right)\\ &=\sum_{m_1=1}^{k}\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}\psi_{m_1}\psi_{m_2}\\ \end{align*}

In the last line we use the identity $\binom{k}{m_1}\binom{m_1}{k-m_2}=\binom{k}{m_2}\binom{m_2}{k-m_1}$ and (21) follows.

Let's summarise:

According to (18) and (19) we get

\begin{align*} \sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m=1}^{j}\binom{j}{m} m!\left(\frac{B_{n,m}^{f}}{f^{m}} + \frac{B_{n,m}^{h}}{h^{m}}\right)\\ &=k!\left(\frac{B_{n,k}^{f}}{f^{k}} + \frac{B_{n,k}^{h}}{h^{k}}\right)\qquad\qquad k \geq 1\tag{34} \end{align*}

and according to (20) and (21) we get

\begin{align*} \sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m_1=1}^j\binom{j}{m_1}m_{1}!\frac{B_{i,m_1}^f}{f^{m_1}}\sum_{m_2=1}^j\binom{j}{m_2}m_{2}!\frac{B_{n-i,m_2}^{h}}{h^{m_2}}\\ &\quad=\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}m_{1}!m_{2}!\frac{B_{i,m_1}^f}{f^{m_1}}\frac{B_{n-i,m_2}^{h}}{h^{m_2}}\qquad\qquad k\geq 1\tag{35}\\ \end{align*}

With the help of (34) and (35) the identity (17) simplifies to \begin{align*} k!\frac{B_{n,k}^{f\cdot h}}{(fh)^{k}}&=k!\left(\frac{B_{n,k}^{f}}{f^{k}} + \frac{B_{n,k}^{h}}{h^{k}}\right)+\sum_{l=1}^{n-1}\binom{n}{l}\\ &\qquad\cdot\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}m_{1}!m_{2}!\frac{B_{l,m_1}^f}{f^{m_1}}\frac{B_{n-l,m_2}^{h}}{h^{m_2}}\\ \end{align*}

We finally get

\begin{align*} B_{n,k}^{f\cdot h}&=h^k B_{n,k}^{f} + f^k B_{n,k}^{h}\\ &\qquad+\frac{1}{k!}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}m_{1}! m_{2}!f^{k-m_1}h^{k-m_2}\\ &\qquad\qquad\cdot \sum_{l=1}^{n-1}\binom{n}{l}B_{l,m_1}^fB_{n-l,m_2}^{h}\tag{36}\\ \end{align*}

Step 2.4: Calculation of $B^{f\cdot (\ln \circ g)}_{n,k}$

Now it's time to harvest the first time. We can now combine the expression with Bell polynomials of $f\cdot h$ and of $(\ln\circ g)$ to calculate the Bell polynomials of $f\cdot (\ln\circ g)$.

We conclude: \begin{align*} B_{n,k}^{f\cdot (\ln \circ g)}&= (\ln \circ g)^k B_{n,k}^{f} + f^k B_{n,k}^{\ln \circ g}\\ &\qquad+\frac{1}{k!}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}m_{1}! m_{2}!f^{k-m_1} (\ln \circ g)^{k-m_2}\\ &\qquad\qquad\cdot\sum_{l=1}^{n-1}\binom{n}{l}B_{l,m_1}^fB_{n-l,m_2}^{\ln \circ g}\tag{37} \end{align*}

According to (14) we know \begin{align*} B^{\ln\circ g}_{n,k}&=\sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j} \end{align*}

Substituting this expression in (37) and summing over $k=1,\ldots,n$ gives:

\begin{align*} \sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln \circ g)} &=\sum_{k=1}^{n}\left((\ln \circ g)^k B_{n,k}^{f} + f^k \sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}\right)\\ &\qquad+\sum_{k=1}^{n}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\frac{m_2!}{(k-m_1)!}\binom{m_1}{k-m_2}f^{k-m_1} (\ln \circ g)^{k-m_2}\\ &\qquad\qquad\cdot\sum_{l=1}^{n-1}\sum_{j=m_2}^{n-l}\frac{(-1)^{j-m_2}}{g^j}\begin{bmatrix}j\\m_2\end{bmatrix}\binom{n}{l}B_{l,m_1}^fB^{g}_{n-l,j}\tag{38} \end{align*}

Summary of part 2: With expression (38) we have developed a formula for $\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln \circ g)}$ and part 2 is finished. The last challenge is to show that OPs expression is equal to this one. This is the theme of part 3.

Note: The body of this answer is limited with $30000$ chars. I'll provide part 3 in a follow-up answer.