$A\in M_n(\mathbb C)$ invertible and non-diagonalizable matrix. Prove $A^{2005}$ is not diagonalizable

If $A^m$ is diagonalizable, then $A^m$ is cancelled by its minimal polynomial $P$, which has simple roots. Therefore $A$ is cancelled by $P(X^m)$ which has simple roots because $P(0)\neq 0$ ($A$ is invertible).

Indeed, if $P(X)=\prod (X-\lambda_i)$, then $P(X^m)=\prod (X^m-\lambda_i)$ whose roots are all the $m$-roots of $\lambda_i$ which differ one frome another (if $\mu_i$ is a $m$-root of $\lambda_i$, then $\mu_i\neq \mu_j$ else $\lambda_i=\mu_i^m=\mu_j^m=\lambda_j$).

As luck would have it, the implication: $A^n$ diagonalizable and invertible $\Rightarrow A$ diagonalizable, was discussed in XKCD forum recently. See my answer there as well as further dicussion in another thread.