If $A=AB-BA$, is $A$ nilpotent? Let matrix $A_{n\times n}$, be such that there exists a matrix $B$ for which
$$AB-BA=A$$
Prove or disprove that there exists $m\in \mathbb N^{+}$such 
$$A^m=0,$$
I know
$$tr(A)=tr(AB)-tr(BA)=0$$
then I can't.Thank you

With the risk of being repetitive, let me just record here the (original?) proof of N. Jacobson in this paper, where you only need to assume that $[A,B]$ and $A$ commute to deduce $[A,B]$ is nilpotent.

Let $[A,B]=A'$, and consider $D(X) = [X,B]$. Then for any polynomial $F$, we have that $D(F(A)) = F'(A)A'$. Now pick a polynomial such that $F(A)=0$ (this can always be done, since the matrices $\{1,\ldots,A^N\}$ are linearly dependent if $N$ is large).
Since $A'$ commutes with $A$, iterating the above we see that if $F$ has degree $d$, then we have that $F^{(d)}(A)A'^{(2d-1)}=d! A'^{(2d-1)}$. Hence, over a field of zero characteristic we get $A'$ nilpotent.

