= Solution
For the positive integer $k$ in the nilpotence condition, diagonalize $B=P\Lambda P^{-1}$. Then
$$
0=B^k=P\operatorname{diag}(\lambda_1^k,\ldots,\lambda_n^k)P^{-1}.
$$
Every 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 is
$$
\boxed{C=\begin{pmatrix}0&1\\0&0\end{pmatrix}\ne0,\qquad C^2=0.}
$$
Its 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>.
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