The Order Of An Inverse

Let $G$ be a group and let $a\in G$ have order $n$. Hence we know that $a^n=e$. Inverting both sides gives us $(a^n)^{-1}=e^{-1}$ which simplifies to $(a^{-1})^n=e$. This shows that the order of $a^{-1}$ is no more than $n$, so we have $|a^{-1}|\leq n$. We must show that if $(a^{-1})^m=e$ for some positive integer $m$, then we must have $m\geq n$. Suppose $(a^{-1})^m=e$ for some positive integer $m$. Then inverting both sides and simplifying gives $(a)^m=e$. Since we know the order of $a$ is $n$, this means that $m\geq n$, as $n$ is the smallest positive integer such that $a^n=e$. This completes the proof that $|a|=|a^{-1}|.$