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? 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?
You should use induction. The base case is $m=1$ and the relation to be proved is
$$
(a^1)^{-1}=(a^{-1})^1
$$
which is true.

Suppose the statement holds for $m$; then
\begin{align}
(a^{m+1})^{-1}
&amp;=(a^ma)^{-1} &amp;&amp;\text{definition of powers}\\
&amp;=(a^m)^{-1}a^{-1} &amp;&amp;\text{because $(xy)^{-1}=x^{-1}y^{-1}$} \\
&amp;=(a^{-1})^ma^{-1} &amp;&amp;\text{induction hypothesis} \\
&amp;=(a^{-1})^{m+1} &amp;&amp;\text{definition of powers}
\end{align}

The proof of $(xy)^{-1}=x^{-1}y^{-1}$:
\begin{align}
(xy)(x^{-1}y^{-1})
&amp;=(yx)(x^{-1}y^{-1}) &amp;&amp;\text{commutativity} \\
&amp;=y\bigl(x(x^{-1}y^{-1})\bigr) &amp;&amp;\text{associativity} \\
&amp;=y\bigl((xx^{-1})y^{-1}\bigr) &amp;&amp;\text{associativity} \\
&amp;=\dots \\
&amp;=1
\end{align}
which you can finish.
$$\forall a,m\in\mathbb{R}$$
Step one: I prove $(a^{m})^{-1}=(a^{-1})^{m}$
$$(x^n)^m=x^{nm}$$
Then
$$(a^{m})^{-1}=(a^{-1})^{m}\Rightarrow a^{m\cdot-1}=a^{-1\cdot m}\Rightarrow a^{-m}=a^{-m}$$

Step two: $a^{-m}&gt;0$
$$a^{-m}=a^{-1\cdot m}=(a^{-1})^{m}=(\frac{1}{a})^{m}=(\frac{1}{a^{m}}) \Longrightarrow(\frac{1}{a^{m}})&gt;0$$