Upper central series members are completely divisibility-closed in nilpotent group

Statement
Suppose $$G$$ is a nilpotent group. Then, the members of the upper central series of $$G$$ (with the possible exception of the zeroth member, the trivial subgroup) are completely divisibility-closed subgroups of $$G$$. Explicitly, this means that if $$p$$ is a prime number such that $$G$$ is $$p$$-divisible, then all the members of the upper central series of $$G$$ (possibly excepting the zeroth member, the trivial subgroup) are completely divisibility-closed subgroups of $$G$$.

In particular, the center and second center are both completely divisibility-closed subgroups, and hence also divisibility-closed subgroups.

Related facts

 * Center not is divisibility-closed

Similar facts

 * Lower central series members are divisibility-closed in nilpotent group

Opposite facts

 * Lower central series members need not be completely divisibility-closed in nilpotent group

Facts similar to proof technique

 * Epicentral series members are completely divisibility-closed in nilpotent group
 * Annihilator of divisibility-closed subgroup under bihomomorphism is completely divisibility-closed
 * Complete divisibility-closedness is strongly intersection-closed

Proof
The proof of this is unusual in that the induction proceeds downward from the penultimate member (the largest). The idea is to shift the "take the $$p^{th}$$ root" operation between the elements in the iterated commutator map and note that the output is unaffected. This forces all $$p^{th}$$ roots of an element in the appropriate upper central series member to be in that upper central series member.