Powering-invariance is commutator-closed in nilpotent group

Statement
Suppose $$G$$ is a nilpotent group and $$H,K$$ are powering-invariant subgroups of $$G$$. Then, the commutator $$[H,K]$$ is also a powering-invariant subgroup of $$G$$.

Related facts

 * Powering-invariance is strongly join-closed in nilpotent group
 * Powering-invariance is not commutator-closed
 * Powering-invariance is not finite-join-closed