Finite implies p-divisible iff p-powered

Statement
Suppose $$G$$ is a finite group and $$p$$ is a prime number. Then, the following are equivalent:


 * 1) $$G$$ is $$p$$-divisible, i.e., every element of $$G$$ has a $$p^{th}$$ root in $$G$$.
 * 2) $$G$$ is $$p$$-powered, i.e., every element of $$G$$ has a unique $$p^{th}$$ root in $$G$$.

Related facts
It is also true that the above equivalent conditions hold if and only if $$p$$ is relatively prime to the order of $$G$$, which in this case means that it does not divide the order of $$G$$ (since it is prime). For a proof, see kth power map is bijective iff k is relatively prime to the order.

(2) implies (1)
This is obvious.

(1) implies (2)
(1) can be reinterpreted as saying that the $$p^{th}$$ power map from $$G$$ to itself is surjective. Since $$G$$ is finite, this implies that the map is bijective, which is equivalent to (2).