For an odd prime $p$, a finite $p$-group is called powerful if $[G,G] \subseteq G^p$, where $G^p = \langle g^p ~:~ g \in G \rangle$.
Prove that : $[G,G] \subseteq [G,G,G] G^p$ implies $G$ is powerful.
The condition implies that $[G,G] G^p = [G,G,G] G^p$ despite $[G,G,G] \subsetneq [G,G]$. This should somehow imply that $[G,G,G] \subseteq G^p$. Trying to commute with $G$ to the set inequality $[G,G] \subseteq [G,G,G] G^p$, and using commutator formulas it only derive $\gamma_3(G) \subseteq \gamma_4(G) \gamma_2(G)^p \gamma_p(G)$. (Here $\gamma_{i+1}(G) := [\gamma_i(G), G]$.)
This makes it further complicated and I can't get rid of the factor $\gamma_p(G)$. Any help?
Do these phenomena happen along with lower central series very often? i.e., if $N \unlhd G$ a proper normal subgroup so that $\gamma_i(G)N = \gamma_{i+1}(G)N$, then is it true that $\gamma_{i+1}(G) \subseteq N$?
Hint: show that $\gamma_2(G)G^p=\gamma_3(G)G^p$ implies $\gamma_2(G)G^p=\gamma_i(G)G^p$ for all $i \gt 1$, by using the general facts that if $H \leq G, N \unlhd G$, then $[G,HN]=[G,H][G,N]$ and $[G,N] \subseteq N$. Finally use that a $p$-group is nilpotent ($\gamma_i(G)=1$ for large enough $i$).