Index of Subgroup in Alternating Group $A_n$

I am trying to show that for $n\geq 5$, the alternating group $A_n$ has no subgroup of index $p$ where $p$ prime and $p\not = n$. I am supposed to show this without using any of the Sylow theorems.

I understand that $A_n$ is simple for $n\geq5$ but I do not know how to relate this to the existence of subgroups of index $p$. I tried to show this by contradiction, assuming $H \subseteq G$ with $[G:H]=p$ and I managed to show that if this is the case, then $p<n$ but I cannot figure out anything beyond that.