Group theory exercise: proof that $b^2 = e$ if $|a|$ is odd and $b = a^{-1}b^{-1}a$ 
Please could someone help me with the last step in this exercise?

I've been stuck. The exercise is this:
Let $a,b$ be two elements in a group such that $a$ has odd order and $b = a^{-1}b^{-1}a$. Show that $b^2 = e$.
What I tried:
First of all let's assume that $|a|=2n + 1$ for some $n$. To show the claim one can  show $b=b^{-1}$. Let's try this:
$$ b = a^{(2n + 1)}ba^{-(2n + 1)} = a^{2n}b^{-1}a^{-2n} = a^{-1}b^{-1}a$$
But this is just the same as the assumption. 
I feel like I'm this close but I'm missing something. 
 A: Note that $$b^{-1}=(a^{-1}b^{-1}a)^{-1}=a^{-1}ba.$$
You should continue your argument, killing one $a$ at a time:
$$\begin{aligned}b&=a^{(-2n+1)}ba^{(2n+1)} \\&=a^{-2n}b^{-1}a^{2n}\\&=a^{-(2n-1)}ba^{(2n-1)}\\
&=\cdots\\
&=a^{-0}b^{-1}a^{0}=b^{-1}.\end{aligned}$$

Another argument based on automorphism group:
The conjugation $c_a(g)=a^{-1}ga$ is an automorphism of $\langle b\rangle$. Obviously $(c_a)^{2n+1}=c_{a^{2n+1}}=id$, where $2n+1$ is the order of $a$. 
On the other hand $c_a(b)=b^{-1}$, so $(c_a)^2$ is the identity on $\langle b\rangle$. 
Thus, on $\langle b\rangle$, $c_a=(c_a)^{2n+1}$ is also the identity homeomorphism.
A: Rewrite the relation as $a=bab$.  Substitute this into itself and you get $a=b^kab^k=bab$ for all $k\ge 1$.
Now, suppose $|a|=2n+1$ and $|b|=m$.  We have
$$e=a^{2n+1}=\underbrace{(bab)(bab)\cdots (bab)}_{2n+1}=$$
$$=(bab)(b^{m-1}ab^{m-1})(bab)(b^{m-1}ab^{m-1})(bab)\cdots (b^{m-1}ab^{m-1})(bab)=ba^{2n+1}b=b^2$$
Note that we need $|a|$ to be odd because it is only every other $bab$ term we are turning into $b^{m-1}ab^{m-1}$.
A: First notice that : $b = a^{-1}b^{-1}a \Leftrightarrow a = bab .$
Thus
$$ba = babb^{-1} = ab^{-1}.$$
and 
$$ ab = b^{-1}bab = b^{-1}a = a^{-1}ab^{-1}a = a^{-1}baa = a^{-1}ba^2.$$
Let $|a| = 2n+1$, then
$$ b = a^{2n+1} b = a^{2n}(ab) = a^{2n}a^{-1}ba^2 = a^{2n-1}ba^2 = \cdots =  aba^{2n} = b^{-1}baba^{2n} = b^{-1}a^{2n+1} = b^{-1}$$
which means 
$$ b^2 = e .$$

A simpler way:
First notice that : $b = a^{-1}b^{-1}a \Leftrightarrow a = bab .$
Thus 
$$ a^2 b^{-1} = a(bab)b^{-1} = aba = b^{-1}baba = b^{-1}(bab)a = b^{-1}a^2. $$
So $b^{-1}$ commute with $a^2$, which means that $b$ commutes with $a^{-2}$ thus all $a^{-2k}$ with $k\in \mathbb{N}$. 
Let $|a| = 2n + 1$, then $a = a^{-2n}$. So $b$ commutes with $a$. So
$$a = bab = ab^2 $$
and finally 
$$ b^2 = e.$$
A: $b=a^{-1}b^{-1}a$ (i.e. $a^{-1}ba=b^{-1}$), conjugate by $a$ both sides: 
$$a^{-1}ba=a^{-2}b^{-1}a^2 \Longrightarrow b^{-1}=a^{-2}b^{-1}a^2.$$
So $a^2$ commutes with $b^{-1}$, hence $\langle a^2\rangle=\langle a\rangle$ commutes with $b^{-1}$ (since $2\nmid o(a)$).
Then what does this implies if $a$ commutes with $b^{-1}$
 in your question?
