Suppose $A^2B+BA^2=2ABA$.Prove that there exists a positive integer $k$ such that $(AB-BA)^k=0$. Let $A, B \in M_n(\mathbb{C})$ be two $n \times n$ matrices such that 
$$A^2B+BA^2=2ABA$$
Prove that there exists a positive integer $k$ such that $(AB-BA)^k=0$.
Here is the source of the problem. Under the comment by Carcul, I don't understand why $[A,[A,B]]=0$ implies that $[A,B]$ is a nilpotent matrix. 
Can anyone explain it to me?
Clarification: I have problem understanding the solution given by Carcul. I don't see the link between $[A,[A,B]]=0$ and $[A,B]$ being nilpotent. 

Due to the assumptions
$$
(AB-BA)^{k+1} = (AB-BA)(AB-BA)^k = AB(AB-BA)^k - B(AB-BA)^kA.
$$
Then we find for the trace
$$
tr( (AB-BA)^{k+1} ) = tr \left( A \big(B(AB-BA)^k\big) - \big(B(AB-BA)^k\big)A\right)=0,
$$
since $tr(CD)=tr(DC)$. This shows that
$$
tr((AB-BA)^{k})=0
$$
for all $k\ge 1$. This implies that $AB-BA$ is nilpotent:
Traces of all positive powers of a matrix are zero implies it is nilpotent
