Say $A$ is a complex $n\times n$ matrix. $A$ is invertible, and moreover, supposer that the set $\{||A^n||:n=1,2,...\}$ is bounded. Prove that $A$ is diagonalizable.
My attempt was to do the following. Say $J$ is the Jordan Canonical form of $A$. Since norm is invariant under similar matrices, we have that $J^n$ is bounded as well. Let $B_1,..,B_k$ be the Jordan blocks appearing in $J$. Assume for sake of contradiction that there is a $j$ such that $B_j$ has size bigger than $1$. Then
$B_j=\lambda I +S$, where $S$ is the matrix with $1$'s in the entries above the diagonal and $0$ elsewhere.
I also know that $\lambda\neq 0$, but I dont know how to finish the problem. Any thoughts?