Anton’s congruence

For every $n\\in\\mathbb{N}$ $\\left(n\\underline{!}\\right)_{p}$ stands for the product of numbersbetween $1$ and $n$ which are not divisible by a given prime $p$ . And we set $\\left(0\\underline{!}\\right)_{p}=1$ .

The corollary below generalizes a result first found by Anton, Stickelberger,and Hensel:

Let $N_{0}$ be the least non-negative residue of $n\\pmod{p^{s}}$ where $p$ is aprime number and $n\\in\\mathbb{N}$ . Then

\\left(n\\underline{!}\\right)_{p}\\equiv\\left(\\pm 1\\right)^{\\left\\lfloor n/p^{s}%\\right\\rfloor}\\cdot\\left(N_{0}\\underline{!}\\right)_{p}\\pmod{p^{s}}.

Proof.

We write each $r$ in the product below as $ip^{s}+j$ to get

$\\displaystyle\\left(n\\underline{!}\\right)_{p}$	$\\displaystyle=$	$\\displaystyle\\prod\\limits_{\\begin{subarray}{c}1\\leq r\\leq n\\\\p^{s}\ot\\div r\\end{subarray}}r$
	$\\displaystyle=$	$\\displaystyle\\left(\\prod\\limits_{\\begin{subarray}{c}0\\leq i\\leq\\left\\lfloor n/%p^{s}\\right\\rfloor-1\\\\1\\leq j<p^{s}\\\\p^{s}\ot\\div j\\end{subarray}}ip^{s}+j\\right)\\left(\\prod\\limits_{%\\begin{subarray}{c}i=\\left\\lfloor n/p^{s}\\right\\rfloor\\\\1\\leq j\\leq N_{0}\\\\p^{s}\ot\\div j\\end{subarray}}ip^{s}+j\\right)$
	$\\displaystyle\\equiv$	$\\displaystyle\\prod\\limits_{i=0}^{\\left\\lfloor n/p^{s}\\right\\rfloor-1}\\prod%\\limits_{\\begin{subarray}{c}1\\leq j<p^{s}\\\\p^{s}\ot\\div j\\end{subarray}}j\\cdot\\prod\\limits_{\\begin{subarray}{c}j=1\\\\p^{s}\ot\\div j\\end{subarray}}^{N_{0}}j)$
	$\\displaystyle\\equiv$	$\\displaystyle\\left(p^{s}\\underline{!}\\right)_{p}^{\\left\\lfloor n/p^{s}\\right%\\rfloor}\\cdot\\left(N_{0}\\underline{!}\\right)_{p}\\pmod{p^{s}}.$

From Wilson’s theorem for prime powers it follows that

\\left(n\\underline{!}\\right)_{p}\\equiv\\begin{cases}\\left(N_{0}\\underline{!}%\\right)_{p}\\text{if}&p=2,s\\geq 3\\\\(-1)^{\\left\\lfloor n/p^{s}\\right\\rfloor}\\cdot\\left(N_{0}\\underline{!}\\right)_{%p}&\\text{otherwise.}\\end{cases}\\pmod{p^{s}}.

∎