Show that $ e^{A+B}=e^A e^B$ If $A$ and $B$ are $n\times n$ matrices such that $AB = BA$ (that is, $A$ and $B$ commute), show that 

$$ e^{A+B}=e^A e^B$$ 

Note that $A$ and $B$ do NOT have to be diagonalizable.
 A: $$\begin{align*}e^{A}e^{B} &= \left(\sum \frac{A^{n}}{n!}\right)\left(\sum\frac{B^{n}}{n!}\right)\\
          &=\sum^{\infty}_{m=0}\sum^{\infty}_{n=0}\frac{A^{m}B^{n}}{m!n!}\\
          &=\sum^{\infty}_{l=0}\sum^{l}_{m=0}\frac{A^{m}B^{l-m}}{m!(l-m)!}\\
          &=\sum^{\infty}_{l=0}\frac{1}{l!}\sum^{l}_{m=0}\frac{l!}{m!(l-m)!}A^{m}B^{l-m}\\
          &=\sum^{\infty}_{l=0}\frac{(A+B)^{l}}{l!}\\
          &= e^{A+B}\end{align*}$$
Note:A and B have to commute.
Also, I set l=m+n.  I did this because we want to use the binomial theorem to simplify this.  
A: Here is a different-ial way, just because it is significantly different from the standard Cauchy product way.
Given a square matrix $M$, the function $X(t):=e^{tM}$ is the unique solution of the linear differential equation: $X'=MX$ and $X(0)=I$.
Now set $X(t):=e^{tA}e^{tB}$ and observe that the factors commute with each other, as well as they commute with $A$ and $B$. It follows that
$$
X'(t)=Ae^{tA}e^{tB}+e^{tA}Be^{tB}=(A+B)e^{tA}e^{tB}=(A+B)X(t).
$$
And since $X(0)=e^0e^0=I$, it follows from the uniqueness above that
$$
X(t)=e^{tA}e^{tB}=e^{t(A+B)}\qquad\forall t\in\mathbb{R}.
$$
Set $t:=1$ to get the desired formula.
