Bourbaki

Proposition 2

Tag 0A5M, permanent. Cite it as [Bourbaki, Tag 0A5M].

Lie Groups and Lie Algebras, chapter II, § 5. Magnus algebras

Suppose that X is finite. For every integer n \geqslant 1, the group F(X)/Fn^{(p)}(X) is a finite p-group of nilpotency class \leqslant n.*

Arguing by induction on nn, it suffices to prove that Fn(p)(X)/Fn+1(p)(X)F_n^{(p)}(X)/F_{n+1}^{(p)}(X) is a finite commutative pp-group for all n⩾1n \geqslant 1. For all w∈Fn(p)(X)w \in F_n^{(p)}(X), the element γ(w)−1\gamma(w) - 1 of A^(X)\hat{A}(X) is of order ⩾n\geqslant n; we denote by δn(w)\delta_n(w) the homogeneous component of γ(w)−1\gamma(w) - 1 of degree nn. The mapping δn:Fn(p)(X)→An(X)\delta_n : F_n^{(p)}(X) \to A^n(X) is a homomorphism with kernel Fn+1(p)(X)F_{n+1}^{(p)}(X) (§ 4, no. 5, Proposition 3) and hence Fn(p)(X)/Fn+1(p)(X)F_n^{(p)}(X)/F_{n+1}^{(p)}(X) is isomorphic to a subgroup of An(X)A^n(X). Since XX is finite, An(X)A^n(X) is a finite-dimensional vector space over Fp\mathbf{F}_p and hence a finite commutative pp-group and so is Fn(p)(X)/Fn+1(p)(X)F_n^{(p)}(X)/F_{n+1}^{(p)}(X).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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