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For the positive integer $k$ in the nilpotence condition, diagonalize $B=P\Lambda P^{-1}$. ThenEvery diagonal entry must vanish, so each ${\lambda }_j=0$. Consequently $\Lambda =0$ and $\boxed{B=0}$. This proves that a diagonalizable nilpotent matrix is zero.

Without diagonalizability a counterexample isIts only eigenvalue is zero but its kernel has dimension one, so it cannot have a basis of eigenvectors. It is the simplest nontrivial nilpotent Jordan block.

Write $B=QD{Q}^{-1}$ with $D$ a diagonal matrix. Then $0=B^k=Q{D}^k{Q}^{-1}$, so each diagonal entry ${\lambda }_i$ satisfies ${\lambda }_i^k=0$ and therefore ${\lambda }_i=0$. Hence $D=0$ and $B=0$. Conversely the zero matrix is diagonalizable and nilpotent. This proves the existing result that a diagonalizable nilpotent matrix is zero; nilpotence alone does not imply a zero matrix, because nonzero nilpotent Jordan blocks exist.