Let $P$ be a $p$ -group and $\Phi(P)$ its Frattini subgroup.

Every maximal subgroup $Q$ of $P$ is of index $p$ in $P$ and is therefore normal in $P$ . Thus $P/Q\cong \mathbb{Z}_p$ . So given $g\in P$ , $g^p\in Q$ which proves $P^p\leq Q$ . Likewise, $\mathbb{Z}_p$ is abelian so $[P,P]\leq Q$ . As $Q$ is any maximal subgroup, it follows $[P,P]$ and $P^p$ lie in $\Phi(P)$ .

Now both $[P,P]$ and $P^p$ are characteristic subgroups of $P$ so in particular $F =[P,P]P^p$ is normal in $P$ . If we pass to $V=P/F$ we find that $V$ is abelian and every element has order $p$ - that is, $V$ is a vector space over $\mathbb{Z}_p$ . So the maximal subgroups of $P$ are in a 1-1 correspondence with the hyperplanes of $V$ . As the intersection of all hyperplanes of a vector space is the origin, it follows the intersection of all maximal subgroups of $P$ is $F$ . That is, $[P,P]P^p=\Phi(P)$ .