Commutator with exponential $[\exp(A),\exp(B)]$ $A,B$ are quantum mechanical operators. 


*

*$[A,B]\neq 0$ that is given.

*$e^{A}=\sum_{n=1}^{\infty} \frac{A^n}{n!} $
Is the following correct?
$$[e^{A},e^{B}]=e^{A}e^{B}-e^{B}e^{A}=e^{A+B}-e^{B+A}=0 $$
 A: An even simpler example:
\begin{align}
s_z=\left(
\begin{array}{cc}
 1 & 0 \\
 0 & -1 \\
\end{array}
\right)\, ,& \qquad s_y=\left(
\begin{array}{cc}
 0 & i \\
 -i & 0 \\
\end{array}
\right)\\
e^{-i\alpha s_z}=\left(
\begin{array}{cc}
 e^{-i \alpha } & 0 \\
 0 & e^{i \alpha } \\
\end{array}
\right)&\qquad e^{-i\beta s_y}=\left(
\begin{array}{cc}
 \cos (\beta ) & \sin (\beta ) \\
 -\sin (\beta ) & \cos (\beta ) \\
\end{array}
\right)
\end{align}
and clearly $[e^{-i\alpha s_z} ,e^{-i\beta s_y}]\ne 0$.
Note that, in connection to the comments on the answer of @user124864,
$e^{-i\alpha s_z}e^{-i\beta s_y}\ne e^{-i\alpha s_z-i\beta s_y-\frac{1}{2}\alpha\beta[s_z,s_y]}$ either.
You can easily verify using the first few terms of the explicit expansion that, in general 
$$
e^A e^B= \sum_n \frac{A^n}{n!}\,\sum_m \frac{B^m}{m!}
\ne e^{A+B}=\sum_{p}\frac{(A+B)^p}{p!}\, .
$$
If anything:
\begin{align}
e^Ae^B &=\left(1+A+\frac{A^2}{2!}+\ldots \right)
\left(1+B+\frac{B^2}{2!}+\ldots \right)\\
&= 1+ (A+B)+ \frac{1}{2!} (A^2+2AB+B^2)+\ldots \tag{1}
\end{align}
but 
$$
(A^2+2AB+B^2)\ne (A+B)^2= A^2+AB+BA+B^2 \tag{2}
$$
unless $AB=BA$, i.e. unless $[A,B]=0$.
A: No this is not correct. The problem is that you have used $e^Ae^B = a^{A+B}$. This holds when $A$ and $B$ are numbers, so you might expect that it would also holds for matrices or even general operators. This is not in general the case, as the other answers show explicitly.
