If $(G,*)$ is a group then for all $a, b \in G$, we have $(a*b)'=b'*a'$. Which would be the analogous expression for $(a*b'*b)'$? Here's the full problem:

Show through calculations and the theorem $2.3$ that if $G$ is a group with binary operation $*$ then for all $a, b \in G$, we have $(a*b)'=b'*a'$. Which would be the analogous expression for $(a*b'*b)'$?

The theorem 2.3 that I'm refering to, is the existance of $a$ unique module ($e*a=a*e=a$) and inverse ($a'*a=a*a'=e$).
I have doubts in the process that I'm attempting to do. Here's what I got so far:
Given $a,b \in G$
$a*b$
$(a*b)$; associative (1)
$(a*b)\in G$
$(a*b)'*(a*b)=(a*b)*(a*b)'=e$; T2.3
$(a*b)*(a*b)'=e$
$(a*a'*b*b')*(a*b)'=e*a'*b'$; T2.3 (2)
$e*e*(a*b)'=a'*b'$
$(a*b)'=a'*b'$
$(a*b)'=b'*a'$; commutative (?) (3)
I'm starting my studies with group theory, so I'm not sure if I'm able to do what I did in (1), (2) and (3). Any help in finding a different way to solve this and the final question?

(NB: Not every group is commutative. Consider the dihedral group $D_3$ for example.)

We have
$$\begin{align}
\color{red}{(ab)}\color{blue}{(b^{-1}a^{-1})}&=a((bb^{-1})a^{-1})\\
&=a(ea^{-1})\\
&=aa^{-1}\\
&=e
\end{align}$$
and, similarly, $\color{blue}{(b^{-1}a^{-1})}\color{red}{(ab)}=e$; thus your theorem gives that $\color{red}{(ab)}^{-1}=\color{blue}{b^{-1}a^{-1}}$.

Since $b^{-1}b=e$, we get $(ab^{-1}b)^{-1}=(ae)^{-1}=a^{-1}.$
