Prime divisibility of Stirling numbers of first kind

Question

Prove that $p\mid\genfrac[]0{}{p^w}k$ where $p$ is an odd prime, $w \in \mathbb{N}$, $1<k<p^w$ and $k \neq p^v$ for some positive integer $v<w$. This has to be already done I just can't find where. Where in the literature has this been stated?

Answer 1

(The $w=1$ case is well-known, so we suppose $w\ge2$ in what follows; we also let $p$ denote any prime, which need not be odd.) See the remarks following the proof of Theorem 2.2 in Fredric T. Howard, “Congruences for the Stirling numbers and associated Stirling numbers”, Acta Arithmetica 55 (1990), 29–41. In particular, it is stated that ${hp+m\brack k}\equiv0\pmod p$ if $m\in\{0,1,2,p-2,p-1\}$ and $h\ge1$ (with a few exceptions given); set $h=p^{w-1}$ and $m=0$, so that the only applicable exception is $${hp\brack h+(p-1)i}\equiv{h\choose i}(-1)^{h-i}\pmod p$$ for $0\le i\le h$. Then, since $p$ divides ${h\choose i}$ for $0<i<h$, we deduce that $${p^w\brack k}\equiv0\pmod p,$$ except for the two cases $${p^w\brack p^{w-1}}\equiv-1 \quad\text{and}\quad {p^w\brack p^w}\equiv1$$ (modulo $p$).

Interestingly, an analogous result for Stirling numbers of the second kind is also given later in the paper (following Theorem 4.2): For $t>0$, we have $$\begin{align*} {p^t\brace k}\equiv0\pmod p\qquad&\hbox{if $k\ne p^r$, $0\le r\le t$,}\\ {p^t\brace p^r}\equiv1\pmod p\qquad&(r=0,1,\dots,t). \end{align*}$$