 Power lifting in a powerful p-group

For odd $p$, set $P_0=G$ and $P_{i+1}=P_i^p$. Powerful embedding gives $[P_i,G]\le P_{i+1}$. Modulo $P_{i+2}$, commutators with $P_i$ are central and have exponent dividing $p$, so $(xy{)}^p\equiv x^p{y}^p$ when $y\in P_i$. Also the power map $P_i/P_{i+1}\to P_{i+1}/P_{i+2}$ is a surjective homomorphism. First find a root modulo $P_2$, then multiply it by successive corrections in $P_i$. The process terminates for a finite p-group, and converges for a pro-p group with separated power filtration. This proves that the generated power subgroup consists of actual powers.