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Let $F^{(r)}$ denote the free group of rank $r$ with generators $x_1, \dots, x_r$.

Recall a group $G$ is nilpotent of index $s$ if $G_{s+1} = \{e\} $ and $G_s \neq \{e\}$ (where $G_i$ denotes the lower central series of $G$)

We may then define $ G = N_{r,s} = F^{(r)}/F^{(r)}_{s+1}$ to be free nilpotent group of index $s$ and rank $r$ with generators $u_i = x_iF^{(r)}_{s+1}$.

Now let $S = \{1 , u_1^{\pm 1}, \dots, u_r^{\pm 1}\}$.

I am asked to show that $|S^n| \leq n^{O_{r,s}(1)}|S|$, where $O_{r,s}(1)$ means some constant that is allowed to depend only on $r$ and $s$.

Essentially I am asked to show that $G$ has polynomial growth of degree $O_{r,s}(1)$, and I am given a finite, symmetric generating set for $G$ containing $1$.

My issue though, is that I am aware that if we are instead considering the growth of $F^{(r)}$ using the "same" generating set (or rather, the appropriate lift) then the growth rate is exponential, (I believe we would have something like $|S'^n| = \frac{r}{r-1}(2r-1)^n - \frac{1}{r-1}$).

The fact then that adding in a nilpotency conditions drops us from exponential growth to polynomial is telling me that the proportion of words that are trivial for large $n$, because of the nilpotence, is large. However, I'm not sure how to write that down rigorously, and more importantly I'm not sure how to actually show that this proportion is large, or how it would tell me that the growth rate is not polynomial.

I am not even sure how to begin thinking about this question to be honest, because I'm not sure how to accurately "categorise" all words that contain trivial commutators and I'm not sure how I can effectively count them. Since I know the explicit size of $|S'^n|$, then if I were able to count them I believe I would be done, but I would have to effectively show that there are exponentially many of them.

I would really appreciate if someone could let me know if I'm barking up the wrong tree here, and I'd be grateful for any help you may be able to offer, thank you.

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