linear algebra - Proving that $(A^n)^{-1} = (A^{-1})^n$ for invertible matrix $A$.

I have seen a proof of the fact that for an invertible matrix $A$, $A^n$ is also invertible and

$$(A^n)^{-1} = (A^{-1})^n.$$

The proof was by induction and it was mentioned that one has to use induction because one has that it is true for all $n$.

I am wondering why one has to use induction. Why can't one just say that

$$(A^n)(A^{-1})^n = (AA^{-1})^n = I^n =I$$
where one has used that $A$ and $A^{-1}$ by definition commute. Isn't this enough to show that $A^n$ is also invertible and $(A^n)^{-1} = (A^{-1})^n$?

Answer

your proof is fine (to me, at least). I think if you wanted to be super rigorous, you'd need to use induction anyways to fully show that $(A^n)(A^{-1})^n=(AA^{-1})^n$ because even though $A$ and $A^{-1}$ commute, you need to be sure it holds for arbitrarily large products of the two.

Honestly though, like I said, your proof is fine how it is.