Finite implies powering-invariant

Statement
Suppose $$G$$ is a group and $$H$$ is a finite subgroup of $$G$$. Then, $$H$$ is a powering-invariant subgroup of $$G$$. Explicitly, if $$p$$ is a prime number such that $$G$$ is powered over $$p$$, then $$H$$ is also powered over $$p$$.

Related facts

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

Proof
Given: A group $$G$$, a finite subgroup $$H$$, a prime number $$p$$ such that the map $$x \mapsto x^p$$ is bijective from $$G$$ to itself.

To prove: The map $$x \mapsto x^p$$, restricted to $$H$$, is bijective from $$H$$ to itself.

Proof: