Commutator-verbal implies divisibility-closed in nilpotent group

Statement
Suppose $$G$$ is a nilpotent group and $$H$$ is a commutator-verbal subgroup of $$G$$. Then, $$H$$ is a divisibility-closed subgroup of $$G$$. Explicitly, for any prime number $$p$$ such that $$G$$ is a $$p$$-divisible, $$H$$ is also $$p$$-divisible.

In particular, this also shows that $$H$$ is a powering-invariant subgroup of $$G$$.