So the question asks: Give a full proof, by mathematical induction, that if the two n × n matrices A and B are similar, then so are $A^n$ and $B^n$, for every $n ∈ N.$
So so far I have:
Base case: A and B are similar, $B=S^{-1}AS$ where S is an invertible matrix
Inductive step: assume $B^n = (S^{-1}AS)^n = (S^{-1}AS) * (S^{-1}AS) * (S^{-1}AS)...(S^{-1}AS) $by n times =$ (S^{-1}A^tS)$
Now multiply both sides by B get,
$B^n * B = (S^{-1}A^tS) * B, $
$B^{n+1} = (S^{-1}A^tS) * S^{-1}AS = S^{-1}A^{t+1}S $
So $B^{n+1} = S^{-1}A^{t+1}S $
Hence for n=n+1 it is true. So it is true for all n. Statement is proved by induction.
Does this look right?
Yes, your induction is correct. Two matrix $A$ and $B$ are similar if exist a invertible matrix $M$ such that $$A=M^{-1}BM$$ Multiplying $n-1$ times for $A$ both sides it's easy to see that $$A^n=M^{-1}B^nM$$ because $M^{-1}M=I$