Prove or disprove $\exp(AB)=\exp(BA)$ where $A$ and $B$ are square matrices 
Is there a necessary and sufficient condition for 
  exp(AB) = exp(BA) where $A,B\in\mathbb{C}^{n\times n}$

Notice that AB does not have to equal to BA.
What I can prove now is that det(exp(AB))=det(exp(BA)) since tr(AB)=tr(AB) implies exp(tr(AB))=exp(tr(BA)), which is sufficient that det(exp(AB)) = det(exp(BA)). Note that exp(tr(A))=det(exp(A)). This is a corollary of Jacobi's formula.
Thanks a lot!!


*

*Jacobi's formula 

*On the Necessary and Sufficient Condition for a Set of Matrices to Commute and Some Further Linked Results
 A: Consider a nonsingular matrix $X$.  It will have various logarithms (i.e. matrices whose exponential is $X$): let two of them be $Y$ and $Z$, and suppose $Y$ and $Z$ are similar.  Thus there is a 
nonsingular matrix $A$ such that $Z = A^{-1} Y A$.  Take $B = A^{-1} Y$, so
$Z = BA$ and $Y = AB$, and $\exp(AB) = \exp(BA)$.
Conversely, any case where $A$ is invertible will arise in this way, taking
$X = \exp(AB)$, $Y = AB$ and $Z = BA$.
A: Of course, finding a NSC valid for any $n$ is hopeless. Yet, the case $n=2$ is relatively simple. 
Let $A,B\in M_2(\mathbb{C})$ s.t. $AB\not= BA$. It can be proved that
$e^{AB}=e^{BA}$ IFF 
EDIT. EITHER i) $A$ or $B$ is invertible, for example $A$
$A$ is invertible and $B=A^{-1}P$ where $P$ is similar to $diag(u,u+2ki\pi)$ with $u\in\mathbb{C},k\in\mathbb{Z}\setminus\{0\}$.
OR ii) $A$ and $B$ are singular
Up to a change of basis, $A=\begin{pmatrix}0&0\\a_{2,1}&a_{2,2}\end{pmatrix},B=\begin{pmatrix}0&b_{1,2}\\0&b_{2,2}\end{pmatrix}$, where
$a_{2,1}b_{1,2}+a_{2,2}b_{2,2}=2ki\pi$ with $k\in\mathbb{Z}\setminus\{0\}$.
$\textbf{Remark}$. In both cases, $e^{AB}$ is a scalar matrix.
$\textbf{Proposition}$. When $n\geq 3$, there are couples $(A,B)$ s.t. $e^{AB}=e^{BA},AB\not= BA$ and $e^{AB}$ is not diagonalizable.
$\textbf{Proof}$. Choose, for $n=3$,
$A=\begin{pmatrix}0&-2&0\\0&0&0\\0&12i\pi&6i\pi\end{pmatrix},B=\begin{pmatrix}-1/2&0&-1\\0&-1/2&0\\0&1&1/3\end{pmatrix}$. 
We obtain $e^{AB}=\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}$.
