$M,N\in \Bbb R ^{n\times n}$, show that $e^{(M+N)} = e^{M}e^N$ given $MN=NM$
Another take on it, which avoids the somewhat tedious term-by-term manipulation and term-by-term comparison of matrix power series:
Consider the ordinary, constant coefficient, matrix differential equation
$dX / dt = (M + N)X; \, \text{with} \, X(0) = I; \tag{1}$
the unique matrix solution is well-known to be
$X(t) = e^{(M + N)t}. \tag{2}$
Next, set
$Y(t) = e^{Mt}e^{Nt} \tag{3}$
and note that, by the Leibnitz rule for derivatives of products,
$dY / dt = (d(e^{Mt}) / dt) e^{Nt} + e^{Mt}(d(e^{Nt}) /dt) = Me^{Mt}e^{Nt} + e^{Mt}Ne^{Nt}, \tag{4}$
and since $MN - NM = [M, N] = 0$ we also have $[e^{Mt}, N] = 0$ so that (4) becomes
$dY / dt = Me^{Mt}e^{Nt} + Ne^{Mt}e^{Nt} = (M + N)e^{Mt}e^{Nt} = (M + N)Y, \tag{5}$
and evidently
$Y(0) = I, \tag{6}$
so that $X(t)$ and $Yt)$ satisfy the same differential equation with the same initial conditions; thus $X(t) = Y(t)$ for all $t$, or
$e^{(M + N)t} = e^{Mt}e^{Nt} \tag{7}$
for all $t$. Taking $t = 1$ yields the requisite result.QED
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \begin{align} \color{#0000ff}{\large\expo{M}\expo{N}} &= \sum_{\ell = 0}^{\infty}{M^{\ell} \over \ell!} \sum_{\ell' = 0}^{\infty}{N^{\ell'} \over \ell'!} = \sum_{\ell = 0}^{\infty}\sum_{\ell' = 0}^{\infty}{M^{\ell}N^{\ell'} \over \ell!\ell'!} \sum_{n = 0}^{\infty}\delta_{n, \ell + \ell'} \\[3mm]&= \sum_{n = 0}^{\infty}\sum_{\ell = 0}^{\infty}{M^{\ell} \over \ell!} \sum_{\ell' = 0}^{\infty}{N^{\ell'} \over \ell'!}\,\delta_{\ell',n - \ell} = \sum_{n = 0}^{\infty} \sum_{\ell = 0 \atop {\vphantom{\LARGE A}n - \ell\ \geq\ 0}}^{\infty} {M^{\ell} \over \ell!}\,{N^{n - \ell} \over \pars{n - \ell}!} \\[3mm]&= \sum_{n = 0}^{\infty}{1 \over n!} \sum_{\ell = 0}^{n} {n! \over \ell!\pars{n - \ell}!}\,M^{\ell}N^{n - \ell} = \sum_{n = 0}^{\infty}{1 \over n!} \sum_{\ell = 0}^{n}{n \choose \ell}M^{\ell}N^{n - \ell} \\[3mm]& =\sum_{n = 0}^{\infty}{1 \over n!}\pars{M + N}^{n} = \color{#0000ff}{\large\expo{M + N}}\,,\qquad\mbox{since}\quad\bracks{M,N} = 0 \end{align}
Hint.
$$(M+N)^2=M^2+MN+NM+N^2=(\text{if}~~[M,N]=0)=M^2+2MN+N^2.$$
Use this fact in the expansion of $\exp(M+N)$ to arrive at sum of monomials of the form $M^qN^r$ with $q,r\geq 0$ and rational coefficients (without $[M,N]=0$ you would have also monomials of the form $N^rM^q$!) . To finish the proof you need to collect the monomials of the same degree in $\exp(M)\exp(N)$, where the ordering is clear by definition.
The product of two series (one of which is absolutely convergent) is $$\left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right)=\sum_{n=0}^\infty \left(\sum_{k=0}^n a_{n-k}b_{k}\right).$$ Applying this to the series, $$e^Me^N = \left(I + M + \frac{M^2}{2!} + \frac{M^3}{3!}\ldots\right) \left(I + N + \frac{N^2}{2!} + \frac{N^3}{3!} \ldots\right)\\=I+ (MI+IN)+\left(\frac{M^2}{2}I+MN+I\frac{N^2}{2}\right)+\ldots$$
Now compare this to the other sum
$$e^{(M+N)} = I+ (M+N) + \frac{(M+N)^2}{2!} + \frac{(M+N)^3}{3!} + \ldots$$