$\mathrm{\Phi }(P)=P^{\mathrm{\prime }}{P}^p$ for a finite $p$-group

Every automorphism sends $g^p$ to $\alpha (g{)}^p$, so it preserves the generating set in [F1] and hence $P^p$ is characteristic; conjugation by an element of $P$ is an automorphism, so $P^p$ is normal. By [L2], $P^{\mathrm{\prime }}$ is normal as well, so [L3] makes $K:=P^{\mathrm{\prime }}{P}^p$ a subgroup, and $gK{g}^{-1}=(g{P}^{\mathrm{\prime }}{g}^{-1})(g{P}^p{g}^{-1})=P^{\mathrm{\prime }}{P}^p=K$ for every $g\in P$ makes it normal.

The elementary abelian quotient $P\mathrm{/}\mathrm{\Phi }(P)$ is abelian and has exponent $p$ by [L1], so it kills every commutator and every $p$th power. Thus $K=P^{\mathrm{\prime }}{P}^p\le \mathrm{\Phi }(P)$.

The quotient $P\mathrm{/}K$ is abelian because it kills $P^{\mathrm{\prime }}$, and every element has $p$th power one because it kills $P^p$. It is therefore elementary abelian, so the kernel criterion in [L1] gives $\mathrm{\Phi }(P)\le K$. Together with step 2.1 this proves equality.