$AB-BA$ is a nilpotent matrix if it commutes with $A$

I use this theorem:

If $\forall i\ge1$, $\mathrm{tr}(C^i)=0$, then $C$ is nilpotent.

You can easily prove by induction that $\mathrm{tr}(C^i)=0$ for all $i\ge1$.

Theorem. $\forall i\ge1$, $\mathrm{tr}(C^i)=0$ iff $C$ is nilpotent.

Proof: $C$ is a real matrix but you may assume $C$ is a complex matrix and $$f(x)=(x-a_1)(x-a_2)\cdots(x-a_n)$$ is its characteristic polynomial in the complex field. You can prove that $\mathrm{tr}(C^k)=\sum_{i=1}^n a_i^k$ by induction, and if $\forall k\in\mathbb N$, $\mathrm{tr}(C^k)=\sum_{i=1}^n a_i^k=0$ then $a_i=0$. Hence, $f(x)=x^n$, so $C^n=0$ and it is shown that $C$ is nilpotent.

Suppose $\mathcal A$ is a normed algebra and $\delta:\mathcal A\to\mathcal A$ is a bounded derivation. If $x\in A$ and $\delta(\delta(x))=0$, then $\lim\limits_{n\to\infty}\|\delta(x)^n\|^{1/n}=0$. This is proved as Theorem 2.2.1 in Sakai's Operator algebras in dynamical systems.

This applies to the case where $\mathcal A=M_n(\mathbb C)$, $\delta(X)= AX-XA$. For an $n$-by-$n$ real or complex matrix $C$, $\lim\limits_{n\to\infty}\|C^n\|^{1/n}=0$ if and only if $C$ is nilpotent.

(I also remarked on this application in an answer to a different question where the result was applicable.)