Every normal subgroup satisfies the quotient-to-subgroup powering-invariance implication

Original formulation
Suppose $$G$$ is a group and $$H$$ is a normal subgroup of $$G$$. Suppose $$p$$ is a prime number such that $$G$$ is powered over $$p$$ (every element of $$G$$ has a unique $$p^{th}$$ root). Suppose that the quotient group $$G/H$$ is also powered over $$p$$. Then, $$H$$ is also powered over $$p$$.

Corollary formulation
Suppose $$G$$ is a group and $$H$$ is a normal subgroup of $$G$$ that is a quotient-powering-invariant subgroup: if $$G$$ is powered over a prime $$p$$ (i.e., every element of $$G$$ has a unique $$p^{th}$$ root), so is the quotient group $$G/H$$.

Then, $$H$$ is also a powering-invariant subgroup of $$H$$: if $$G$$ is powered over a prime $$p$$, so is $$H$$.

Proof idea
The idea is to apply the uniqueness of $$p^{th}$$ roots in the quotient group to the case of the $$p^{th}$$ roots of the identity element of the quotient group.

Proof details for original formulation
Given: A group $$G$$, a normal subgroup $$H$$ of $$G$$. A prime $$p$$ such that both $$G$$ and $$G/H$$ are powered over $$p$$. An element $$g \in H$$.

To prove: There is a unique element $$x \in H$$ such that $$x^p = g$$.

Proof: Let $$\pi:G \to G/H$$ be the quotient map.

Proof for corollary formulation
This is direct from the original formulation once we note that:

$$G$$ is powered over $$p$$ and $$H$$ is quotient-powering-invariant in $$G$$ $$\implies$$ $$G/H$$ is powered over $$p$$.