A linear algebra problem regarding $AB-BA=A$ Let $A$ be a $n \times n$ complex matrix. Show (1) is equivalent to (2)
(1) There exists $B$ such that $AB-BA = A$
(2) $A^n=0$
Furthermore, prove $B^n\neq 0$ in (1) if $A \neq 0$.

Attmept for (1) $\implies$ (2)
$A^2 = A^2B - ABA = ABA-BA^2$. Therefore $A^2 = \frac12 (A^2B - BA^2)$. Inductively $A^n = \frac 1 n (A^nB- BA^n)$. Take submultiplicative norms, we obtain $\| A^n \|\leq \frac 2n \|A^n \| \|B\| $. But here I don't know how to preceed.

Let $ k $ be a positive integer, Given $ A,B\in\mathscr{M}_{n}\left(\mathbb{C}\right) $ such that $ AB-BA=A $, we have for any $ i \in\mathbb{N} :$
$$ A^{k}=A^{k-1-i}\left(AB-BA\right)A^{i}=A^{k-i}BA^{i}-A^{k-1-i}BA^{i+1} $$
Thus : $$ \sum_{i=0}^{k-1}{A^{k}}=\sum_{i=0}^{k-1}{\left(A^{k-i}BA^{i}-A^{k-1-i}BA^{i+1}\right)} $$
Hence, for any $ k\in\mathbb{N} $ : $$ kA^{k}=A^{k}B-BA^{k} $$
Define the following endomorphism : $ \varphi_{B}: \mathscr{M}_{n}\left(\mathbb{C}\right)\rightarrow\mathscr{M}_{n}\left(\mathbb{C}\right),\ X\mapsto XB-BX \cdot $
Suppose $ A $ isn't nilpotent, then $ \mathbb{N}\subset\mathrm{sp}\left(\varphi_{B}\right) $, because every $ k\in\mathbb{N} $ is an eigenvalue of $ \varphi_{B} $ and $ A^{k}\neq O_{n} $ is the associated eigenvector. But that can't be because we're working on $ \mathscr{M}_{n}\left(\mathbb{C}\right) $ and $ \mathrm{dim}\left(\mathscr{M}_{n}\left(\mathbb{C}\right)\right)<+\infty $ means $ \mathrm{sp}\left(\varphi_{B}\right) $ must be finite.
Thus $ A $ is nilpotent.
