I want to show that for two commuting matrices, say $A$ and $B$, we have $e^{A+B}=e^Ae^B.$
So far I have \begin{align*} e^{A+B}&=\sum_{k=0}^\infty \frac{1}{k!}(A+B)^k \\ &=\sum_{k=0}^\infty \frac{1}{k!}\sum_{j=0}^k \binom{k}{j} A^jB^{k-j} \\ &= \sum_{k=0}^\infty \sum_{j=0}^k\frac{A^jB^{k-j}}{(k-j)!j!} \end{align*}
I don't really know where to go now though. What do I need to do next?
In the last expression, permute the sums in order to obtain
$$\sum _{j=0} ^\infty \sum _{k=j} ^\infty \frac {A^j B^{k-j}} {(k-j)! j!}$$
and finally make the change of summation variable $m = k-j$ to obtain
$$\sum _{j=0} ^\infty \sum _{m=0} ^\infty \frac {A^j B^m} {m! j!}$$
which is easily seen to be
$$\sum _{j=0} ^\infty \frac {A^j} {j!} \sum _{m=0} ^\infty \frac {B^m} {m!} = \Bbb e ^A \Bbb e ^B.$$
Alternatively, given that $A$ and $B$ commute, there is a basis in which both are diagonal, and working in this basis makes computations really easy. Try it!