= Solution
Initially $G_1=G$ is <powerfully embedded> in itself. If $G_i$ is <powerfully embedded> in $G$, then $[G_i,G]\subseteq G_i^p$ and consequently $G_{i+1}=G_i^p$. The preliminary result says that $G_i^p$ is again <powerfully embedded> in $G$, completing the induction.
For a finite <p-group> $A$, the <Frattini subgroup> is $\Phi(A)=A^p[A,A]$. Here $[G_i,G_i]\subseteq[G_i,G]\subseteq G_i^p$, so
$$
\boxed{G_{i+1}=G_i^p=\Phi(G_i),\qquad [G_i,G]\subseteq G_{i+1}.}
$$
Since the <Frattini subgroup> of a nontrivial finite <p-group> is proper, this series eventually reaches $1$. Thus $P_j=G^{p^j}$, with iterated power subgroups, terminates.
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