If $e^A$ and $e^B$ commute, do $A$ and $B$ commute? It is known that if two matrices $A,B \in M_n(\mathbb{C})$ commute, then $e^A$ and $e^B$ commute. Is the converse true?

If $e^A$ and $e^B$ commute, do $A$ and $B$ commute?

Edit: Addionally, what happens in $M_n(\mathbb{R})$?
Nota Bene: As a corollary of the counterexamples below, we deduce that if $A$ is not diagonal then $e^A$ may be diagonal.

Another example: 
$$A=\pmatrix{0&-2\pi\\ 2\pi&0}, \textrm{      }B=\pmatrix{0&-2\pi\\49\cdot 2\pi&0}$$
This is a counterexample of the following statement:
$$e^{A+B}=e^Ae^B\Longrightarrow AB=BA$$
since 
$$e^A=e^B=e^{A+B}=I$$ but 
$$AB=4\pi^2 \pmatrix{-49&0\\ 0&-1},$$
$$BA=4\pi^2 \pmatrix{-1&0\\0&-49}.$$
