A diagonalizable nilpotent matrix is zero

If a diagonalizable matrix satisfies $B^k=0$ for a positive integer $k$, write $B=P\mathrm{diag}({\lambda }_j)P^{-1}$. Then every ${\lambda }_j^k=0$, so every eigenvalue is zero and $B=0$. Diagonalizability is essential: the matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ is nonzero but its square is zero. More generally a nonzero nilpotent operator requires a nontrivial Jordan block.