I have been looking at some papers on the "restricted Burnside problem". On page 4 of Vaughan-Lee and Zelmanov's survey, "[Bounds in the restricted Burnside problem](https://doi.org/10.1017/S144678870000121X)", I think they implicitly are invoking the claim that:



> 
> If $G$ is a Lie algebra over $\Bbb{F}\_p$ with nilpotency class $C$, generated by elements $g\_1,\dotsc,g\_m$, then $\lvert G\rvert \le p^{m^C}$.
> 
> 
> 


How does one prove this claim? Is there a nice combinatorial reason for this (I want to guess that this translates to a nice bound on the dimension of $G$)?



Bounding size of group by number of generators, order of elements, and nilpotency class (Restricted Burnside's)
Converting my [comment](https://mathoverflow.net/questions/447519/bounding-size-of-group-by-number-of-generators-order-of-elements-and-nilpotenc#comment1156329_447519) into an answer:


Let $G = G\_1 \ge G\_2 \ge \cdots$ be the lower central series. Then $G\_k/G\_{k+1}$ is spanned by the left-normed commutators $[x\_1, \dotsc, x\_k]$ with $x\_1, \dotsc, x\_k \in \{g\_1, \dotsc, g\_m\}$, of which there are at most $m^{k-1}(m-1)$ nontrivial for $k > 1$ (since $x\_1 \ne x\_2$). Therefore
$$\dim G \le m + \sum\_{k=2}^C m^{k-1}(m-1) = m^C.$$


This bound is not sharp. For the free Lie algebra, $G\_k/G\_{k+1}$ has a basis given by the basic commutators, and the number of these is given by [*Witt's formula*](https://en.wikipedia.org/wiki/Free_Lie_algebra#Universal_enveloping_algebra):
$$\dim G\_k/G\_{k+1} = \frac1k \sum\_{d \mid k} \mu(d) m^{k/d}.$$


