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?



Prime divisibility of Stirling numbers of first kind
(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”](https://eudml.org/doc/206267), *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
$$\begin{align\*}
{p^t\brace k}\equiv0\pmod p\qquad&\hbox{if $k\ne p^r$, $0\le r\le t$,}\\
{p^t\brace p^r}\equiv1\pmod p\qquad&(r=0,1,\dots,t).
\end{align\*}$$


