Powering-invariance is strongly join-closed in nilpotent group

Statement
Suppose $$G$$ is a nilpotent group and $$H_i, i \in I$$ are all powering-invariant subgroups of $$G$$. Then, the join of subgroups $$\langle H_i \rangle_{i \in I}$$ is also a powering-invariant subgroup of $$G$$.

Related facts

 * Powering-invariance is strongly intersection-closed
 * Powering-invariance is not finite-join-closed
 * Divisibility-closedness is strongly join-closed in nilpotent group
 * Divisibility-closedness is not finite-join-closed (the examples for this are solvable, but cannot be nilpotent)
 * Divisibility-closedness is not finite-intersection-closed (there is an abelian example)

Facts used

 * 1) uses::Divisible subset generates divisible subgroup in nilpotent group

Proof
The proof follows from Fact (1): simply take the set-theoretic union of the subgroups as the "divisible subset" for the appropriate set of primes and argue that the subgroup generated by it is also appropriately divisible and hence appropriately powered.