This is my first proof about groups. Please feed back and criticise in every way (including style & language).
Axiom names (see Wikipedia) are italicised. We use $^{-1}$ to denote inverse elements; $e$ denotes the identity element.
Let $(G, \cdot)$ be a group.
By $\textit{identity element}$, $G \ne \emptyset$.
Now, let $a \in G$.
By $\textit{inverse element}$, $a^{-1} \in G$ and $(a^{-1})^{-1} \in G$.
It remains to prove that $(a^{-1})^{-1} = a$.
\begin{equation*}
\begin{split}
a &= a \cdot e && \text{by }\textit{identity element} \\
&= a \cdot \Big(a^{-1} \cdot (a^{-1})^{-1}\Big) && \text{by }\textit{inverse element} \\
&= (a \cdot a^{-1}) \cdot (a^{-1})^{-1} && \text{by }\textit{associativity} \\
&= e \cdot (a^{-1})^{-1} && \text{by }\textit{inverse element} \\
&= (a^{-1})^{-1} && \text{by }\textit{identity element}
\end{split}
\end{equation*}
QED
PS: What I do know is that I have a personal preference for details and explicitness 🙁
Best Answer
First of all, your proof is absolutely fine! There's a more elegant way to prove this fact though. Since inverses are unique in groups, $(a^{-1})^{-1}$ is the unique element of the group satisfying $$ (a^{-1})^{-1} (a^{-1}) = (a^{-1}) (a^{-1})^{-1} = e. $$ On the other hand, we know that $a$ satisfies $$ a (a^{-1}) = (a^{-1}) a = e, $$ so we conclude $a=(a^{-1})^{-1}$ from the uniqueness of inverse elements in a group.