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.

$$
S^{-1} A^{n} S
=
S^{-1} A S S^{-1} A S \cdots S^{-1} A S
=
B^{n}.
$$
Or, proceeding by induction, assume $B^{k} = S^{-1} A^{k} S$, and get
$$
B^{k+1}
=
B^{k} B
=
S^{-1} A^{k} S S^{-1} A S
=
S^{-1} A^{k+1} S.
$$
