Check my proof that $(ab)^{-1} = b^{-1} a^{-1}$ 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!
 A: These questions are standardly done by going straightforward, definition-based. So for the element $ab$ we seek the element $x$ s.t. $abx = xab = e$. Sure. We know such an element is unique (if not - prove this too).
So let's just do it. $ab  b^{-1} a^{-1} = a (b b ^{-1})  a^{-1} = a  e  a^{-1} = a  a^{-1} = e$. That's one direction.
$b^{-1}a^{-1} * ab = b^{-1} (a^{-1}a) b = b^{-1}b = e$
As for your method above - it looks great. Well done.
A: The definition of  inverse is 
a*a-1 = I       (ie a operated with inverse should give identity element)
a-1*a = I      (ie a inverse operated with a should also give identity element)
so here 
$(AB) B^{-1} A^{-1} = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$
$B^{-1} A^{-1} (AB) = B^{-1}(A^{-1}A)B = B^{-1} I B = B^{-1}B = I$
Here when $B^{-1}A^{-1}$ Operated to AB on both sides and in both case it given I (Identity Matrix ).
$X Z  = I$ means that $Z$ is the inverse of $X$ 
Similarly 
If $ABB^{-1}A^{-1}$ is giving identity element then $B^{-1} A^{-1}$ is the inverse of $AB$.
A: Your way is absolutely fine.  As you note, there is in fact an easier way.  It would be enough to show that the element $c$ such that $(ab)c = e$ is in fact $c = b^{-1} a^{-1}$:
$$\begin{align}
(ab)b^{-1} a^{-1} &= a (b b^{-1}) a^{-1}
\\
&= a e a^{-1}
\\
&= a a^{-1}
\\
&= e.
\end{align}$$
A: For a direct proof systematically using the associative property and the fact that $x = xe = xyy^{-1}$ for all $y\in G$  proceed as follows.
\begin{align}
(ab)^{-1}
=
&
(ab)^{-1}e
\\
=
&
(ab)^{-1}[aa^{-1}]
\\
=
&
(ab)^{-1}\big[(ae)a^{-1}\big]
\\
=
&
(ab)^{-1}\Big[\big(a[bb^{-1}]\big)a^{-1}\Big]
\\
=
&
(ab)^{-1}\Big[([ab]b^{-1})a^{-1}\Big]
\\
=
&
(ab)^{-1}\Big[[ab]\big(b^{-1}a^{-1}\big)\Big]
\\
=
&
\Big[(ab)^{-1}[ab]\Big]\big(b^{-1}a^{-1}\big)
\\
=
&
e\big(b^{-1}a^{-1}\big)
\\
=
&
b^{-1}a^{-1}
\end{align}
A: Consider system of linear equations $Cx =b$.
Let $C = AB$.
$\therefore$ $ABx = b$ .............................................................................................................................(1)
If inverse exists, $x = C^{-1}b$ 
$\therefore$ $x=(AB)^{-1}b$ ......................................................................................................................(2)
Now, multiply equation (1) by $(A)^{-1}$,
$\therefore$ $A^{-1}ABx$ = $A^{-1}b$
$\therefore$ $Bx$ = $A^{-1}b$ ...........................................................................................................................(3)
Now, multiply equation (3) by $(B)^{-1}$,
$\therefore$ $B^{-1}Bx$ = $B^{-1}A^{-1}b$
$\therefore$ $x = B^{-1}A^{-1}b$ ......................................................................................................................(4)
Comparing (2) & (4),
$(AB)^{-1} = B^{-1}A^{-1}.$
