$M,N\in \Bbb R ^{n\times n}$, show that $e^{(M+N)} = e^{M}e^N$ given $MN=NM$ I am working on the following problem. Let $e^{Mt} = \sum\limits_{k=0}^{\infty} \frac{M^k t^k}{k!}$ where $M$ is an $n\times n$ matrix. Now prove that
$$e^{(M+N)}   = e^{M}e^N$$
given that $MN=NM$, ie $M$ and $N$ commute.
Now the left hand side of the desired equality is
$$e^{(M+N)} = I+ (M+N) + \frac{(M+N)^2}{2!} + \frac{(M+N)^3}{3!} + \ldots $$ 
On the right hand side of the equation we have
$$e^Me^N = \left(I + M + \frac{M^2}{2!}   + \frac{M^3}{3!}\ldots\right)      \left(I + N + \frac{N^2}{2!}   +   \frac{N^3}{3!} \ldots\right)   $$
Now basically this is as far as I got... I am unsure on how to work out the product of the two infinite sums. Possibly I need to expand the powers on the left hand side expression but I am unsure how to do this in an infinite sum... If anyone could give me an answer or a hint that can help me forward I would greatly appreciate it. Thanks
 A: $\newcommand{\+}{^{\dagger}}%
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\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}
A: 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.
A: 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
$$e^{(M+N)} = I+ (M+N) + \frac{(M+N)^2}{2!} + \frac{(M+N)^3}{3!} + \ldots$$
A: 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!!!
