For matrices
\[A, \: B\]
, \[e^A e^B =e^{A+B}\]
if and only if \[AB=BA\]
\[A+B=B+A \rightarrow e^{A+B}=e^{B+A} \]
Then
\[e^A e^B=e^{A+B}=e^{B+A}=e^Be^A\]
so \[A, \: B\]
commute.Hence if
\[e^A e^B=e^{A+B}\]
then \[A, \: B\]
commute. Now suppose that \[A, \: B\]
commute, so that in particular \[(A+B)^n= {}^n C_k A^{n-k}B^k\]
\[e^A=I+ \frac{A}{1!}+\frac{A^2}{2!}+...+ \frac{A^n}{n!}+...= \sum^{\infty}_{n=0} \frac{A^n}{n!}\]
\[e^B=I+ \frac{B}{1!}+\frac{B^2}{2!}+...+ \frac{B^n}{n!}+...= \sum^{\infty}_{n=0} \frac{B^n}{n!}\]
\[\begin{equation} \begin{aligned} e^Ae^B &= \sum^{\infty}_{n=0} \frac{A^n}{n!} \sum^{\infty}_{m=0} \frac{B^m}{m!} \\ &= \sum^{\infty}_{n=0} \sum^{n}_{m=0} \frac{A^{n-m}}{(n-m)!} \frac{B^m}{m!} \end{aligned} \end{equation}\]
\[\begin{equation} \begin{aligned} e^{A+B} &= I+ \frac{A+B}{1!}+\frac{(A+B)^2}{2!}+...+ \frac{(A+B)^n}{n!}+...\\ &= \sum^{\infty}_{n=0} \frac{(A+B)^n}{n!} \\ &= \sum^{\infty}_{n=0} \sum^{n}_{m=0} \frac{{}^n C_m A^{n-m}B^m}{n!} \\ &= \sum^{\infty}_{n=0} \sum^{n}_{m=0} \frac{A^{n-m}}{(n-m)!} \frac{B^m}{m!} \end{aligned} \end{equation}\]
Hence
\[a^Ae^B=e^{A+B}\]
.