Let $A$ be a $n\times n$ such that $A=PBP^{-1}$ where $B$ is in Jordan normal form with $\lambda_i(k)_j$ Where $i$ is the size, $k$ is the eigenvalue and $j$ the order.
If $A$ was diagonal($i=1$) then $A^n$ in Jordan form has $\lambda_1(k^n)_j$.
If the Jordan form has Jordan blocks bigger then 1, how do we find $A^n$ In Jordan form?
A Jordon block of size $i$ and eigenvalue $k$ has the form $(kI+t)$ where $t^i=0$. Thus $(kI+t)^n$ will just be the truncated binomial expansion:$$\sum_{r=0}^{i-1} {n \choose r} k^{n-r}t^r$$