$P$ and $A$ are square matrices and $P$ is invertible. Prove that $(P^{-1} AP)^{n} = P^{-1}A^{n}P $ The proposition I would like to prove is following:

Proposition. Let $\mathbf{P}$ and $\mathbf{A}$ be square matrices and a matrix $\mathbf{P}$ be invertible. Prove that $({\mathbf{P}^{-1}} \mathbf {AP})^{n} = \mathbf{P}^{-1}\mathbf{A}^{n}\mathbf{P} $

My attempt:
We check two$^{1}$ cases:
$n = 1$ $ ({\mathbf{P}^{-1}} \mathbf {AP})^{1} = {\mathbf{P}^{-1}} \mathbf {AP}$
$n = 2$ $({\mathbf{P}^{-1}} \mathbf {AP})^{2} = {\mathbf{P}^{-1}} \mathbf {AP}{\mathbf{P}^{-1}} \mathbf {AP}= {\mathbf{P}^{-1}}\mathbf {AI} \mathbf {AP} = {\mathbf{P}^{-1}}\mathbf {A}^2 \mathbf {P} $
Suppose it is true for $n = k$, i.e:
$$\tag ! ({\mathbf{P}^{-1}} \mathbf {AP})^{k} = {\mathbf{P}^{-1}} \mathbf {A^{k}P} $$
Now we need to prove that proposition holds for $n = k+1$:
$$ ({\mathbf{P}^{-1}} \mathbf {AP})^{k+1} = {\mathbf{P}^{-1}} \mathbf {AP}({\mathbf{P}^{-1}} \mathbf {AP})^{k}$$
Using result obtained in $(!)$:
$$({\mathbf{P}^{-1}} \mathbf {AP})^{k+1} = {\mathbf{P}^{-1}} \mathbf {AP}{\mathbf{P}^{-1}} \mathbf {A^{k}P} = {\mathbf{P}^{-1}} \mathbf {A^{k+1}P}$$
As desired. $\Box$
Is the proof correct?



*

*Technically speaking (if I understand the definition of the mathematical induction correctly), case $n=1$ is sufficient as the base case. But is it better (makes evident that there is a pattern) if I show proposition holds for $n = 2$ too?


Your proof looks good.
When you're starting your journey towards becoming a mathematician, it can seem natural to give an extra example like your $n=2$ case to provide extra illustration to the reader.  But if you have confidence in your ability to write clear proofs, you'll come to see it as dead weight that is just making your proof longer.  So feel free to find that confidence as soon as possible.  ^_^
