Let $N$ be a group and $S$ a finite, symmetric generating set with the identity.
For $n \in \mathbb N$, we let $S^n = \{s_1\dots s_n\mid s_i \in S\}$
We say $N$ has polynomial growth rate if $\exists c,d > 0 $ such that $|S^n| \leq cn^d$
I am asked to show that if $N$ is nilpotent, then the growth rate is indeed polynomial.
After doing some research, I have found some "old" papers proving this result. However, these proofs are quite long and hard to understand, and so I expect there might be an easier way.
However, I have an issue that I don't know how to start.
I think that by nilpotency of $N$ we may assume that $S$ is closed under commutators, since all commutators are eventually the identity, and this would guarantee that $S$ is still finite.
Also we can utilise the fact that the growth rate of a group is actually independent of the generating set, and so if we can bound the growth of this $S^n$, then we'd be done.
I have thought about perhaps trying to manually compute the size of $S^n$, but there seem to be too many unknowns here for me to be able to do this.
I have also thought about perhaps finding a subgroup that has polynomial growth, since I am aware that this is equivalent, but I don't know much about the structure of subgroups of a nilpotent group, so I'm not entirely sure if this would be a helpful approach.
I would appreciate any helpful hint that may point me in the right direction regarding how I should approach this, thank you!
