Im trying to prove the following identity of the exponential of a matrix: $$ e^{A+B}= e^{A}e^{B} $$ where $A,B \in \mathbb{M}^{n\times n} $ s.t their product commute. Here is my procedure until now: \begin{eqnarray} e^{A+B} &=& \sum_{n=0}^{\infty} \frac{1}{n!} (A+B)^{n} \\ &=& \sum_{n=0}^{\infty} \frac{1}{n!} \left[\sum_{k=0}^{n}\binom{n}{k} A^{k}B^{(n-k)}\right] \\ &=& \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{A^{k}}{k!}\frac{B^{(n-k)}}{(n-k)!} \end{eqnarray} the previous equation can be rewritten as: $$ e^{A+B} = \sum_{0\leq k \leq n\leq\infty} \frac{A^{k}}{k!}\frac{B^{(n-k)}}{(n-k)!} $$ but I am stuck, in how to change the indices of the sum in order to obtain the last equality: $$ e^{A+B}= \sum_{m=0}^{\infty} \sum_{j=0}^{\infty} \frac{A^{m}}{m!}\frac{B^{j}}{j!} = e^{A}e^{B} $$
Any help or advice is well received.