Why is $a_1\cdots a_{2n+1}a_1^{-1}\cdots a_{2n+1}^{-1}$ a product of $n$ commutators? Let $a_1,\ldots,a_{2n+1}$ be elements of a group $G$. Then
$$a_1\cdots a_{2n+1}a_1^{-1}\cdots a_{2n+1}^{-1}$$
is a product of $n$ commutators. 

The case $n=1$ was proven in this question and the proof of the above statement should use that and induction. I'm ashamed to admit that again I have no clue. It's probably quite analogous, but I really can't think of anything but blind guessing. 
So it would be nice if someone could help me out.   
 A: Let $g = a_3a_4 \cdots a_{2n+1}$. Then
$a_1a_2 \cdots a_{2n+1} a_1^{-1} a_2^{-1} \cdots a_{2n+1}^{-1} =
(a_1a_2g a_1^{-1}a_2^{-1} g^{-1})(g a_3^{-1}a_4^{-1} \cdots a_{2n+1}^{-1})$
which is a product of two expressions of the same type, the first with $n=1$ and the second with $n$ one less than the original $n$.
A: Here's a proof that it's the product of $4n-3$ commutators, but that's not what you asked for!
Let $g=a_2a_3\cdots a_{2n}$ and $h=a_2^{-1}a_3^{-1}\cdots a_{2n}^{-1}$. Now rewrite your product 
$$ a_1^{-1}ha_{2n+1}^{-1}a_1ga_{2n+1} $$
as
$$ g^{-1}(ga_1^{-1}ha_{2n+1}^{-1}a_1ga_{2n+1}g^{-1})g.$$
Conjugation doesn't effect being the product of $n$ commutators, so let's focus on
$$ ga_1^{-1}ha_{2n+1}^{-1}a_1ga_{2n+1}g^{-1}.$$
Now note that
$$ [a_1g^{-1},a_{2n+1}h^{-1}] = ga_1^{-1}ha_{2n+1}^{-1}a_1g^{-1}a_{2n+1}h^{-1}.$$
And so we're reduced to writing the difference between the above two products as some combination of commutators.  But the difference between the two is
$$ ha_{2n+1}^{-1}ga_{2n+1}\cdot a_{2n+1}^{-1}ga_{2n+1}g^{-1}.$$
By induction, we're done.
