Solution
= Solution
\b[True.] Since $\alpha$ is invertible, every <eigenvalue> $\lambda$ is nonzero. If $\alpha$ had a nontrivial Jordan block $\lambda I+N$, then for the stated positive integer $N_0$,
$$
(\lambda I+N)^{N_0}
=\lambda^{N_0}I+N_0\lambda^{N_0-1}N+\cdots
$$
would still have a nonzero nilpotent part and could not be diagonalizable. Therefore all Jordan blocks of $\alpha$ have size one. This also proves <diagonalizability inherited from an invertible power>.