On the proportion of elements of prime order in finite symmetric groups

We give a short proof for an explicit upper bound on the proportion of permutations of a given prime order $p$, acting on a finite set of given size $n$, which is sharp for certain $n$ and $p$. Namely, we prove that if $n\equiv k(\mathrm{m}\mathrm{o}\mathrm{d}p)$ with $0\le k\le p-1$, then this proportion is at most $(p\cdot k!{)}^{-1}$ with equality if and only if $p\le n<2n$.