Exercise:
Prove, that for every element $a$ of the group: $(a^m)^{-1}=(a^{-1})^m$, where $m$ is natural.
On the #23 was proven that $(a_1 \centerdot \ldots \centerdot a_n)^{-1} = a_n^{-1} \centerdot \ldots \centerdot a_1^{-1}$ and I believe that #26 directly follows from this, when $a_i=a$ for all $i$ from $1$ to $n$. Would you be so kind to tell me am I right? (Author gives another, not so simple answer.)
Thanks.