## If N is an nxn nilpotent matrix over a field, then Nⁿ = 0

Let $N$ be an $n \times n$ matrix over a field $F$, and suppose $N^k = 0$. Prove that $N^n = 0$.

We saw in this previous exercise that $N$ is similar to a matrix $J$ which has no nonzero entries on or below the main diagonal. (Specifically, $J$ is the Jordan canonical form of $N$.) As we showed in this previous exercise, $J^n = 0$. Since $N = P^{-1}JP$, we have $N^n = 0$.