Quotient-powering-invariance is quotient-transitive

Statement
Suppose $$G$$ is a group and $$H,K$$ are normal subgroups of $$G$$ with $$H$$ contained inside $$K$$. Suppose that $$H$$ is a quotient-powering-invariant subgroup of $$G$$ and $$K/H$$ is a quotient-powering-invariant subgroup of the quotient group $$G/H$$. Then, $$K$$ is a quotient-powering-invariant subgroup of $$G$$.

Related facts

 * Powering-invariance is not quotient-transitive
 * Powering-invariant over quotient-powering-invariant implies powering-invariant

Proof
The proof is fairly straightforward.