This is from an article by Hall from 1961; it's probably one of the most trivial observations in that article, but I can't get the reasoning.
Let G be some group and let A be a normal abelian p-subgroup of G for some prime p. We want to show that $A^p$, the subgroup of p-th powers of elements of A, is contained in the Frattini subgroup of G.
So let M be a maximal subgroup in G which doesn't contain A and denote $N=M\cap A$. I have managed to show that N is normal in G and that A/N is a minimal normal subgroup of G/N. So A/N can't have any nontrivial proper characteristic subgroups. Therefore $A^pN=N$, in which case we are done, or $A^pN=A$. I don't know where to go from here. I suppose the fact that A is a p-group should come into play somewhere, since all of the above works for any exponent n instead of p.