I'm asked to show that $e^{A+B} = e^{A}e^{B}$ for commuting square matrices. This question has been answered before here and the accepted answer is this one. Intuitively I agree with the proof given but I can't prove it, in particular the following identity: $$\sum^{\infty}_{m=0}\sum^{\infty}_{n=0}\frac{A^{m}B^{n}}{m!n!} =\sum^{\infty}_{l=0}\sum^{l}_{m=0}\frac{A^{m}B^{l-m}}{m!(l-m)!}$$
I tried to get an intuition behind it using partial sums but the identity doesn't seem to hold there. Although it seems to make sense that the indexes are eventually the same when $l$ goes to infinity, the statement is still kind of mysterious to me. How does one get that identity?
Consider the first sum with indices $n,m$ and set $l:=n+m$. Instead of summing over both $n$ and $m$, we sum over $l$ and then consider all possible $(n,m)$ with $n+m=l$: $$\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{A^mB^n}{m!n!}=\sum_{l=0}^{\infty}\sum_{n+m=l}\frac{A^mB^n}{m!n!}.$$ Now, $l=n+m$ is equivalent with $n=l-m$ and if $m+n=l$, we know that $0\le m\le l$, so $$\sum_{n+m=l}\frac{A^mB^n}{m!n!}=\sum_{m=0}^l\frac{A^mB^{l-n}}{m!(l-n)!}.$$ Plugging this back in gives the desired identity.