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. 

Let $X = AB - BA$
From the assumptions $A^2B - ABA = ABA - AB^2 \iff A(AB-BA) = (AB-BA)A \iff AX = XA $ 
By induction: $$ AX^k = X^kA \text{ } \forall k \in \mathbb{N} $$
$X^k = X^{k-1}X = X^{k-1}(AB-BA) = X^{k-1}AB - X^{k-1}BA = AX^{k-1}B - X^{k-1}BA \Rightarrow $
$\operatorname{tr}(X^k) = \operatorname{tr}(AX^{k-1}B)- \operatorname{tr}(X^{k-1}BA) = 0 \text{ } \forall k \in \{1,2,\dots,n\}$
One can show that this leads to the nilpotence of $X = AB-BA$.

Why does $\operatorname{tr}(X^k) \text{ } \forall k \in \{1,2,...,n\}$ imply the nilpotence of $X$ ?
Let $\lambda_1, \lambda_2, \dots \lambda_n $ be the eigenvalues of $X$. Then $\operatorname{tr}(X^k) = \sum\limits_{i=1}^n \lambda_i^k $. The coefficients of the characteristic polynomial ( besides the coefficient of $x^n$ ) are the symmetric elementary polynomial in variables $\lambda_1, \lambda_2, \dots \lambda_n$. Newton's Identities say that: 
$$ ke_k = \sum_{i=1}^k(-1)^{i-1} e_{k - i}  p_i, $$
where $e_k$ are the symmetric polynomials and $p_k = \operatorname{tr}(X^k).$
Because $\operatorname{tr}(X^k) = 0$ we can conclude that $e_1 = e_2 = \dots = e_n = 0$  so the characteristic polynomial of $X$ is $X^n$. By Cayley Hamilton $X^n = 0$. 
