Does $\sum_{i=0}^\infty{i\left((1-p^{i+1})^m-(1-p^{i})^m\right)}$ go to infinity as $\log m$?

Let's denote your sum by $S(m,p)$. First of all, we have $$S(m,p)=\sum_{k=1}^{\infty}\big(1-(1-p^k)^m\big)=\color{blue}{\sum_{j=1}^{m}(-1)^{j-1}\binom{m}{j}\frac{p^j}{1-p^j}}.$$ (The first equality is obtained, with $a_k=1-(1-p^k)^m$, from $$\sum_{k=1}^{n}k(a_k-a_{k+1})=\sum_{k=1}^{n}ka_k-\sum_{k=1}^{n+1}(k-1)a_k=\sum_{k=1}^{n}a_k-na_{n+1};$$ to get the second, expand $(1-p^k)^m$ and sum over $k$ first).

A "tool" for this kind of sums is Nørlund–Rice integral (easily verified by oneself): $$S(m,p)=-\frac{m!}{2\pi i}\oint_{C}F(m,p,z)\,dz,\qquad F(m,p,z)=\frac{(p^{z}-1)^{-1}}{\prod_{j=0}^{m}(z+j)}$$ where $C$ is a closed contour encircling $z=-1,\ldots,-m$ and no other poles of the integrand.

Now, if $C_n$ is the circle $|z|=-(2n+1)\pi/\ln p$ oriented counterclockwise, then $$0=\lim_{n\to\infty}\frac{m!}{2\pi i}\oint_{C_n}F(m,p,z)\,dz=-S(m,p)+m!\sum_{n\in\mathbb{Z}}\operatorname*{Res}_{z=2n\pi i/\ln p}F(m,p,z);$$ evaluation of the residues gives (not just an asymptotic - an exact one!) $$S(m,p)=-\frac{1}{2}-\frac{H_m}{\ln p}+\frac{1}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\Im\prod_{j=1}^{m}\Big(1+\frac{2n\pi i}{j\ln p}\Big)^{-1},$$ where $H_m=\sum\limits_{j=1}^{m}\dfrac{1}{j}$ is the $m$-th harmonic number.

This confirms $\color{blue}{\dfrac{S(m,p)}{\ln m}\underset{m\to\infty}{\longrightarrow}-\dfrac{1}{\ln p}}$, as $H_m=\ln m+O(1)$ when $m\to\infty$.