Positive power of an invertible matrix with complex entries is diagonalizable only if the matrix itself is diagonalizable. Show that a positive power of an invertible matrix with complex entries is diagonalizable only if the matrix itself is diagonalizable. 
The other direction is trivial. This direction seems a little more involved. I'm not sure if it's best to start by assuming $A$ is not diagonalizable and get that $A^k$ isn't as well. Or to start from diagonalization of $A^k$ and work one out for $A$. 
First method: $A$ has non trivial Jordan block. Powers of this block are upper triangular Toeplitz matrices. This means we have generalized eigenvectors and thus we don't have diagonalizability.
Second method: If $A^k$ is diagonalizable, consider it's minimal polynomial, $m$, i.e. the monic polynomial of smallest degree such that $m(A^k)=0$. We have that $m$ factors linearly with no repeated roots. Invertibility gives that none of the eigenvalues can be zero. I'm not sure where to go from here...
 A: It looks like your first method of proof works perfectly well.  However, just for the fun of it, let's approach this from the minimal polynomial perspective.
Suppose that $A^k$ is diagonalizable.  It follows that $A^k$ satisfies its minimal polynomial
$$
p(x) = \prod_{\ell = 1}^m (x - \lambda_\ell)
$$
with each $\lambda_\ell$ distinct.
That is, for $p$ as above, $p(A^k) = 0$.  

First, we consider the case in which $|\lambda_\ell|$ are all distinct. Now, we note that
$$
x^k - \lambda_\ell = \prod_{j=0}^{k-1} (x - \omega^j\sqrt[k]{\lambda_\ell})
$$
where $\omega = e^{2\pi i/k}$ is the usual $k$th root of $1$.  Defining $q(x) = p(x^k)$, we note that $q(A) = 0$, so that the minimal polynomial of $A$ must divide $q$.  However, we can write
$$
q(x) = 
p(x^k) = \prod_{\ell = 1}^m (x^k - \lambda_\ell) = 
\prod_{\ell = 1}^m
\prod_{j=0}^{k-1} (x - \omega^j\sqrt[k]{\lambda_\ell})
$$
That is, $q$ is the product of distinct linear factors.  Thus $A$ would be diagonalizable.

Next, we consider $A^k$, restricted to any span of eigenspaces so that each $|\lambda_\ell|$ is the same.  That is, without loss of generality, we have
$$
S^{-1}A^kS = k \pmatrix{\lambda_1\\&\ddots \\ && \lambda_n}
$$
for some constant $k > 0$ and $|\lambda_\ell| = 1$.  It suffices to prove that, in this case, $A$ is diagonalizable.
