Divisibility is quotient-closed

Statement
Suppose $$G$$ is a group and $$H$$ is a normal subgroup of $$G$$, so that $$G/H$$ is the quotient group. Then, if $$n$$ is a natural number such that $$G$$ is $$n$$-divisible (i.e., every element of $$G$$ has a $$n^{th}$$ root in $$G$$, then so is $$G/H$$, i.e., every element of $$G/H$$ has a $$n^{th}$$ root in $$H$$.

Proof
Given: A group $$G$$ and a natural number $$n$$ such that every element of $$G$$ has a $$n^{th}$$ root in $$G$$, a normal subgroup $$H$$ with quotient group $$G/H$$. An element $$a \in G/H$$.

To prove: There exists $$b \in G/H$$ such that $$b^n = a$$.

Proof: Let $$\varphi:G \to G/H$$ be the quotient map and let $$g \in G$$ be such that $$\varphi(g) = a$$. Since $$G$$ is $$n$$-divisible, there exists $$h \in G$$ such that $$h^n = g$$. Let $$b = \varphi(h)$$. Then, $$b^n = (\varphi(h))^n = \varphi(h^n) = \varphi(g) = a$$, so $$b$$ is as desired.