As the title says, I need to prove that every element in $G/\Phi(G)$ has order $p$. I know by the First Sylow theorem that $G$ contains subgroups of orders $1,p,p^2,\dots,p^{k-1}$ with each normal in the next.
Thus, every maximal subgroup of $G$ has order $p^{k-1}$, so $\Phi(G)$ is the intersection of subgroups of $G$ of order $p^{k-1}$. However from here I'm stuck. I do also know that $\Phi(G)$ itself is contained in a maximal subgroup of $G$ (by Sylow), and thus must either be of order $p^{k-1}$ (in which case $G/\Phi(G) \cong \mathcal{C}_p$ and I'm done) or it's properly contained in a maximal subgroup.
I'm unsure of what to do in the latter case. Any help would be appreciated!