Prove that A(AB-BA) = (AB-BA)A implies AB-BA is nilpotent.

Let A and B be $n \times n$ complex matrices such that $A(AB-BA) = (AB-BA)A$

a) Show that for every positive integer $k$, the matrix $(AB-BA)^k$ is of the form $AC-CA$, where $C$ is an $n \times n$ complex matrix.

b) Prove that $AB-BA$ is nilpotent.

I have tried the following. $A^{-1}A(AB-BA) = A^{-1}(AB-BA)A \implies AB-BA = A^{-1}(AB-BA)A$
$A(AB-BA)A^{-1} = (AB-BA)AA^{-1} \implies AB-BA = A(AB-BA)A^{-1}$
Then $(AB-BA)^{k} = (A^{-1}(AB-BA)A)^k = A^{-1}(AB-BA)^kA$
and $(AB-BA)^{k} = A(AB-BA)^{k}A^{-1}$.
I don't know where do I go from here, thanks for your help.

a) We prove this by induction. For $k=1$, just take $B=C$. Suppose that you can find a matrix $C_k$ such that $(AB-BA)^k=AC_k-C_kA$. Then you have $$(AB-BA)^{k+1} = (AC_k-C_kA)(AB-BA) = A(C_k(AB-BA)) - C_kA(AB-BA), $$ but, by hypothesis you know that $$A(AB-BA) = (AB-BA)A $$ Consequently, $$(AB-BA)^{k+1} = A(C_k(AB-BA)) - (C_k(AB-BA))A, $$ so you can take $C_{k+1} = C_k(AB-BA)$ to get that $(AB-BA)^{k+1} = AC_{k+1}-C_{k+1}A$.

b) Try to use the result proved here.