Endomorphism kernel implies quotient-powering-invariant

Statement
Suppose $$G$$ is a group and $$H$$ is an endomorphism kernel in $$G$$, i.e., $$H$$ is a normal subgroup of $$G$$ and there is a subgroup $$K$$ of $$G$$ such that $$G/H \cong K$$. Then, $$H$$ is a quotient-powering-invariant subgroup of $$G$$, i.e., for any prime number $$p$$ such that $$G$$ is $$p$$-powered, so is the quotient group $$G/H$$.

Related facts

 * Finite normal implies quotient-powering-invariant
 * Normal of finite index implies quotient-powering-invariant

Proof
Given: A group $$G$$, a normal subgroup $$H$$ of $$G$$ such that $$G/H$$ is isomorphic to a subgroup $$K$$ of $$G$$. $$G$$ is powered over a prime number $$p$$.

To prove: $$G/H$$ is powered over $$p$$.

Proof: