Show that a matrix is nilpotent.
Solution 1:
Let's denote by $C$ the matrix $AB-BA$.
From this is enough to show that $\operatorname{Tr}(C^n)=0$ for all $n\in\mathbb N$.
From Mariano's comment we easily see that $AC=CA\Rightarrow AC^2=CAC=C^2A\Rightarrow\ldots \Rightarrow AC^m=C^mA, \ \forall m\in\mathbb N$. Therefore
$$
C^n=C^{n-1}C=C^{n-1}(AB-BA)=\\
AC^{n-1}B-C^{n-1}BA=A(C^{n-1}B)-(C^{n-1}B)A.
$$
We conclude that $\operatorname{Tr}(C^n)=0$ for all $n\in\mathbb N$.