Iterated commutator map is endomorphism for nilpotent group

Statement
Suppose $$G$$ is a nilpotent group and its nilpotency class is $$c$$. Consider the iterated commutator map:

$$T: G^c = G \times G \times \dots \times G \to G$$

given by:

$$T(x_1,x_2,\dots,x_c) = [[ \dots [[x_1,x_2],x_3], \dots, x_{c-1}],x_c]$$

Then, $$T$$ is a multihomomorphism. Explicitly, choose any $$i \in \{ 1,2,\dots,c \}$$. Fix all the values of $$x_j, j \ne i$$. Then, $$T$$, viewed as a function solely of $$x_i$$, defines an endomorphism of $$G$$. In fact, $$T$$ descends to a set map:

$$G/G' \times G/G' \times \dots \times G/G' \to \gamma_c(G)$$

that is a homomorphism in each coordinate.

Related facts

 * Class two implies commutator map is endomorphism
 * Commutator map is homomorphism if commutator is in centralizer

Facts used

 * 1) uses::Commutator map is homomorphism if commutator is in centralizer

Proof outline
We prove the claim by induction, and using Fact (1).

Suppose $$G$$ is a group of nilpotency class $$c$$. By the inductive hypothesis, the $$(c - 1)$$-fold iterated commutator is a homomorphism:

$$T_{c-1}:G/\gamma_c(G) \times G/\gamma_c(G) \times \dots \times G/\gamma_c(G) \to \gamma_{c-1}(G)/\gamma_c(G)$$

In fact, again by the inductive hypothesis, it descends to the following, which is a homomorphism in each coordinate:

$$\overline{T}_{c-1}:G/G' \times G/G' \times \dots \times G/G' \to \gamma_{c-1}(G)/\gamma_c(G)$$

By Fact (1), the commutator map:

$$\gamma_{c-1}(G) \times G \to \gamma_c(G)$$

is a homomorphism in each coordinate. Further, it descends to a homomorphism:

$$\gamma_{c-1}(G)/\gamma_c(G) \times G/\gamma_c(G) \to \gamma_c(G)$$

Combining, we obtain that $$T$$ descends to the following, a homomorphism in each coordinate:

$$T: G/G' \times G/G' \times \dots \times G/G' \times G/\gamma_c(G) \to \gamma_c(G)$$

Finally, since the map is a homomorphism in the last coordinate for each choice of the other coordinates, it descends to a map from $$G/G'$$ in the last coordinate as well, and we get the following as a homomorphism in each coordinate:

$$T: G/G' \times G/G' \times \dots \times G/G' \times G/G' \to \gamma_c(G)$$

Note that the formulation outlined here avoid explicit use of the three subgroup lemma, but that is an alternative way of explaining why the mapping descends.