 Solution

For a powerful p-group, $G_2=G^p$, and $G_3=(G^p{)}^p[G^p,G]$. In $G/G_3$, the image of $G_2$ is central and has exponent dividing $p$, while the commutator subgroup is contained in that image. The quotient thus has nilpotency class at most two. Direct collection in such a group givesbecause $p$ is odd. If $z\in G_2$, the same calculation gives $(xz{)}^p\equiv x^p$ modulo $G_3$, since $z^p\in G_3$. Consequently the power map is well defined on $G/G_2$ and is a group homomorphism to $G_2/G_3$. Its image contains all the power generators of that quotient. It is therefore surjective.