 Solution

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 $N_0$,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.