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. 

From $[A,[A,B]] = 0$, we decuce by induction that 

$$ [A^k, B] = kA^{k-1}[A,B] \tag 1 $$

For $k = 0$, (1) is obviuos, if (1) holds for $k-1$, we have
\begin{align*}
   [A^k, B] &= A^kB - BA^k\\
            &= A^{k-1}AB - BAA^{k-1}\\
            &= A^{k-1}[A,B] + A^{k-1}BA - BA^{k-1}A\\
            &= A^{k-1}[A,B] + [A^{k-1},B]A\\
            &= A^{k-1}[A,B] + (k-1)A^{k-2}[A,B]A\\
            &= kA^{k-1}[A,B] \qquad \text{    as $[A,B]A = A[A,B]$}
\end{align*}
As $[\cdot, B]$, is linear, we have that for any polynomial $p$ we have
$$ [p(A), B] = p'(A)[A,B] $$
Now let $\mu_A$ denote the minimal polynomial of $A$, we will show by induction that 

$$ \tag 2 \mu^{(k)}_A(A)[A,B]^{2^k- 1} = 0 $$

holds for any $k$.
For $k = 0$, we have nothing to show, as $\mu_A(A) = 0$ by definition of the minimal polynomial. Suppose $k \ge 1$ and (2) holds for $k-1$, then
\begin{align*}
  0 &= [\mu^{(k-1)}_A(A)[A,B]^{2^{k-1}-1}, B]\\
    &= [\mu^{(k-1)}_A(A), B][A,B]^{2^{k-1}-1} + \mu_A^{(k-1)}(A)[[A,B]^{2^k-1}, B]\\
    &= \mu^{(k)}_A(A)[A,B]^{2^{k-1}} + \mu_A^{(k-1)}(A)\sum_{l=1}^{2^{k-1}-1}
          [A,B]^{l-1} [[A,B],B][A,B]^{2^{k-1}-l}\\
\end{align*}
As $[A,[A,B]] = 0$, any polynomial in $A$ commutes with $[A,B]$. Multiplying the last equation with $[A,B]^{2^{k-1}-1}$ from the left, we have
\begin{align*}
  0 &= [A,B]^{2^{k-1}-1}\mu^{(k)}_A(A)[A,B]^{2^{k-1}} + [A,B]^{2^{k-1}-1}\mu_A^{(k-1)}(A)\sum_{l=1}^{2^{k-1}-1}
          [A,B]^{l-1} [[A,B],B][A,B]^{2^{k-1}-l}\\
    &= \mu^{(k)}_A(A)[A,B]^{2^k-1} + \sum_{l=1}^{2^{k-1}-1}\mu_A^{(k-1)}(A)
          [A,B]^{2^k-1 + l-1} [[A,B],B][A,B]^{2^{k-1}-l}\\
    &= \mu^{(k)}_A(A)[A,B]^{2^k-1} \quad\text{by induction hypothesis}
\end{align*}
This proves (2). 
Now, in (2), let $k = \deg \mu_A$ to get 
$$ \tag 3 (\deg \mu_A)![A,B]^{2^{\deg \mu_A} - 1} = 0 $$
Hence, $[A,B]$ is nilpotent.
