The following question is problem Pinter's Abstract Algebra. And to put things in context: $G$ is a group and $a, b$ are elements of $G$.
I want to show $(ab)^{-1}$ = $b^{-1}a^{-1}$.
I originally thought of proving the fact in the following manner: \begin{align*} (ab)^{-1}(ab) &= e \newline (ab)^{-1}(ab)(b^{-1}) &= (e)(b^{-1}) \newline (ab)^{-1}(a)(bb^{-1}) &= (b^{-1}) \newline (ab)^{-1}(a)(e) &= (b^{-1}) \newline (ab)^{-1}(a) &= (b^{-1}) \newline (ab)^{-1}(a)(a^{-1}) &= (b^{-1})(a^{-1}) \newline (ab)^{-1}(e) &= (b^{-1})(a^{-1}) \newline (ab)^{-1} &= (b^{-1})(a^{-1}) \newline \end{align*}
I know this may seem extremely inefficient to most, and I know there is a shorter way. But would this be considered a legitimate proof?
Thanks in advance!