Congruence for Stirling Number of first kind $s(n,k)$ when $n$ is prime

Question

Let $s(n,k)$ be the Stirling numbers of first kind: $$\prod_{k=0}^{k=n-1}(x-k) =\sum_{k=0}^{k=n}s(n,k)x^k$$

$p$ is prime $\iff$ for all $k\in\{2,..,p-1\}$, $s(p,k)\equiv0\ mod\ p $

How can this be demonstrated?

There is an analogous statement for the Stirling numbers of second kind ${n \brace k}$ which is quite easy to prove from the explicit formula $$k!{n \brace k}=(-1)^k\sum_{j=0}^{j=k}(-1)^j\binom{k}{j}j^n $$ But I do not know of an analogous explicit formula for the first kind...

Answer 1

Work in the polynomial ring $(\Bbb Z/p\Bbb Z)[x]$: in that ring $x^{\underline p}=x^p-x$, since both polynomials are monic of degree $p$ and have roots $0,\ldots,p-1$. Now just equate coefficients to get the positive part of the result.

For the negative result, suppose that $n$ is not prime, let $p$ be a prime dividing $n$, and let $m=\frac{n}p$. Then

$$\begin{align*} x^{\underline n}&=\prod_{i=0}^{n-1}(x-i)=\prod_{k=0}^{m-1}\prod_{\ell=0}^{p-1}(x-kp-\ell)=\prod_{k=0}^{m-1}\prod_{\ell=0}^{p-1}(x-\ell)\\\\ &=\left(x^{\underline p}\right)^m=\left(x^p-x\right)^m=x^m\left(x^{p-1}-1\right)^m\;, \end{align*}$$

and it follows that $s(n,m)\equiv(-1)^m\not\equiv 0\pmod p$.

For many more results on congruences involving Stirling numbers of the first kind see Rhodes Peele, A.J. Radcliffe, and Herbert S. Wilf, Congruence problems involving Stirling numbers of the first kind [PDF]. What I did here is a special case of results there.

Answer 2

Here is another very different argument, which seems to work but only in one direction:$$x^{\underline n}=\prod_{i=0}^{n-1}(x-i)=(-1)^nn!\sum_{\sum_{i=1}^{i=n} {i.n_i=n}}\prod_{i=1}^{i=n}\frac{(-x)^{n_i}}{i^{n_i}n_i!}$$ The big sigma is over all the integer partitions of $n$, i.e. over all the n-uplets $(n_1,..,n_n), n_i\ge0$ such that $\sum_{i=1}^{i=n} {i.n_i=n}$

then $$s(n,k)=(-1)^{n+k}n!\sum\prod_{i=1}^{i=n}\frac{1}{i^{n_i}n_i!}$$ here the sum is over all the integer partitions of $n$ with $k$ summands.

Since $n_i\le n$, all the prime divisors of $\prod_{i=1}^{i=n}i^{n_i}n_i!$ are $\le n$, but actually apart from the single partition with $k=n$ summands $(n_1=n,n_i=0, i\gt1)$ and the single partition with $k=1$ summand $(n_i=0, i \lt n,n_n=1 )$ all the prime divisors of $\prod_{i=1}^{i=n}i^{n_i}n_i!$ are $\lt n$ because in these cases $n_n=0$ and $n_i\lt n$. Then, the denominator on the right hand side of the above explicit formula for $s(n,k)$ has only prime factors which are $\lt n$, then if $n$ is prime, it is also a divisor of $s(n,k)$.

From here I don't know how to complete the demonstration in the other direction..