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](https://mathoverflow.net/questions/15316/collapsible-group-words/) 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.



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


