Proving that if matrix A is similar to B, A^n is similar to B^n. Matrix A is similar to matrix B <=> B=S$^-$$^1$AS
Prove that if A is similar to matrix B => A$^n$ is also similar to to B$^n$. (n is a member of ℤ).
Without thinking too much, the first idea that came into my mind was proof by induction.
for n=1: B=S$^-$$^1$AS
for n=k: B$^k$=S$^-$$^1$A$^k$S
for n=k+1: B$^k$$^+$$^1$=S$^-$$^1$A$^k$$^+$$^1$S
B$^k$B=S$^-$$^1$A$^k$AS
B$^k$B=S$^-$$^1$A$^k$SS$^-$$^1$AS 
I'm not sure how to decompose the proof at this point. I could say that due to our inductive assumption, B$^k$ = S$^-$$^1$A$^k$S is sure to be a true statement. What is left is B=S$^-$$^1$AS, which is also true given the very first assumption. But I don't think this is the right way to decompose the proof. Can anyone help me?
Thank you in advance.

This is a standard manipulation.  Think about going in both directions in this equation: $$\begin{align}
    B \cdot B = (S^{-1}AS) \cdot (S^{-1}AS) = S^{-1}A(S S^{-1})AS = S^{-1}A^2 S
\end{align}$$
