Let $G$ be a finite group of order $p^e$ for some prime $p$. Let $S$ be a set of size not divisible by $p$. I know that
$|G| = |Stab(s)|\cdot|O_s|$ = (stabilizer of $s$)(orbit of $s$)
So if there is no fixed point then the $|O_s|$ is never $1$, so by the class equation
$|G| = \sum_{orbits} |O_s| = \sum_{orbits} p^{e(s)}$, since $|O_s|$ divides $|G|$. Since there is no $1$ in the sum, it's divisible by some maximal $p^\ell$. Factoring out $p^\ell$, then we have $|G|/p^\ell = p^k = 1 + \dots $ (multiples of $p$), which is impossible, since $1$ is not a multiple of $p$.
Am I missing any thing to make this a complete proof?
Two criticisms:
It is perfectly possible to divide $|G|$ by $p^\ell$ and get $$|G|/p^\ell=\underbrace{1+1+\cdots+1}_{{\rm multiple~of~}p}+({\rm multiples~of~}p).$$
In any case the class equation yields the size of $S$ not of $G$ so the above is moot. You could still divide $|S|$ by the maximal $p^\ell$ and get multiple $1$s in the resulting sum. The point is not the number of $1$s on the RHS after dividing by the $p^\ell$, but the fact that $p$ divides $|S|$ at all!
The idea here is: no fixed point $\implies |S|=\sum ({\rm multiples~of~}p)\implies 0\not\equiv |S|\equiv0$ mod $p$, absurd.