Product of nilpotent matrices is nilpotent

Question

Let $A$ and $B$ be two nilpotent $n\times n$ matrices that commute (so $AB=BA$), how do I show that $AB$ is nilpotent as well?

I have frankly no idea how to start this proof, so excuse me for not showing what I have done so far.

Answer 1

Since A,B commute you know that $ (AB)^n = A^nB^n$ A is nilpotent, so there exists $ m \in \mathbb N$ such that $A^m=0$. so $\:(AB)^m=0B^m=0$