If $G$ is a finite group, then if $n$ is the number of elements of order $p$, $p\mid n+1$ when $n\neq 0$.
Suppose $n\neq 0$. Since there is an element of order $p$, $p$ divides the order of the group, so we can take a maximal abelian $p$-group $H$. Let $P$ denote the set of all elements of order $P$, and let $H$ act on $P$ by conjugation. The orbits must divide the order of $H$, and partition $P$. So modulo $p$, $n\equiv f\pmod{p}$ where $f$ is the number of fixed points.
Of course, $H$ fixes all its elements of order $p$, (which form a subgroup when the identity is adjointed), so it has $p^m-1$ elements in $P$ for some $m$.
By maximality, it follows somehow that $H$ fixes only those $p^m-1$ elements in $P$. Why can't it fix more? If it did fix more, does that contradict maximality?
After that it follows easily that $n\equiv f=p^m-1\equiv-1\pmod{p}$. Thank you for your time.