Maximal power in a sequence of iterated commutators in the rank two free group

Question

I have the following problem: in the free group $F_2=\langle a,b\rangle$, we define the sequence

$\begin{cases} w_0=a, \\ w_1=b, \\ w_{n+2}=[w_{n+1},w_{n}] & \text{for }n\ge 0. \end{cases}$

So $w_2$ is the classical commutator $[b, a]$ (I take $[a,b]=aba^{-1}b^{-1}$ but this doesn't really matter), and then you keep iterating it.

We find $w_3=bab^{-1}a^{-1}baba^{-1}b^{-2}$, and so on. One remarks that the maximal exponent in the reduced expression of $w_3$ is $2$ (for the last $b^{-1}$). This property seems to hold for every $w_n$. I have checked this on my computer up to the word $w_{20}$. I am looking for a (nice?) proof!

As a weaker problem, I would be happy with a proof that exponents are uniformly bounded.

This one seems to be a close question, but I do not see how estimates of powers may come into play.

Clearly, I have been thinking of proving it by induction. One problem is that the length of $w_{n+1}$ is nearly the double of $w_n$ (just an experimental observation), meaning that when writing $w_{n+2}=[w_{n+1},w_n]=w_{n+1}w_nw_{n+1}^{-1}w_n^{-1}$ and then making the simplifications, a third of the word gets lost in simplifications.

Answer 1

A word in $F_2$ can be represented by a path on the unit square grid on the plane. Now, $w_0$ is a horizontal unit interval, $w_1$ is a vertical unit interval, $w_2$ is a unit square and the image of $w_3$ is a union of two adjacent squares (on on the top of the other; it looks like a figure 8 from an old calculator). Observe that both $w_2^{\pm 1}$ and $w_3^{\pm 1} $ are closed loops based at the origin both contained in the rectangle $[0,1]\times [0,2]$. It follows that $w_4$ is also a loop based at the origin and contained in this rectangle. And hence every $w_n$ has this property. It follows that the reduced expression of $w_n$ has maximal exponent $2$ and, moreover, this is the exponent of $b^{\pm 1}$ only.