How do you prove$(a^m)^{-1} = (a^{-1})^m$ for a being any non-zero real and m being a positive integer from the axioms of real numbers?

Question

How do you prove $(a^m)^{-1} = (a^{-1})^m$ for $a$ being any non-zero real and $m$ being a positive integer, from the axioms of real numbers only?

Answer 1

From the existence of the inverse we have that, for $a \ne0$ there exists $b$ such that $a^mb=1$.

For $m=1$ this gives $ab=1 \Rightarrow b=a^{-1}$ .

Now suppose that $a^mb^m=1$ than: $$ a^{m+1}b^{m+1}=a^mab^mb=a^mb^mab=a^mb^m=1 $$

and this prove, by induction, the statement.