$A\in M_n(\mathbb C)$ invertible and non-diagonalizable matrix. Prove $A^{2005}$ is not diagonalizable $A\in M_n(\mathbb C)$ invertible and non-diagonalizable matrix. I need to prove that $A^{2005}$ is not diagonalizable as well. I am asked as well if Is it true also for  $A\in M_n(\mathbb R)$. (clearly a question from 2005).
This is what I did: If $A\in M_n(\mathbb C)$ is invertible so $0$ is not an eigenvalue, We can look on its Jordan form, Since we under $\mathbb C$, and it is nilpotent for sure since $0$ is not an eigenvalue, and it has at least one 1 in it's semi-diagonal. Let $P$ be the matrix with Jordan base, so $P^{-1}AP=J$ and $P^{-1}A^{2005}P$ but it leads me nowhere.
I tried to suppose that $A^{2005}$ is diagonalizable and than we have this $P^{-1}A^{2005}P=D$
When D is diagonal and we can take 2005th root out of each eigenvalue, but how can I show that this is what A after being diagonalizable suppose to look like, for as contradiction? 
Thanks

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.
