Wilson’s theorem for prime powers

For every natural number $n$, let ${\left(n\underset{¯}{!}\right)}_p$ denote the product of numbers $1\le m\le n$ with $gcd(m,p)=1$.

For prime $p$ and $s\in \mathbb{N}$

Proof: We pair up all factors of the product ${\left(p^s\underset{¯}{!}\right)}_p$ into thosenumbers $m$ where $m\not\equiv m^{-1}(mod{p}^s)$ and those where this is notthe case. So ${\left(p^s\underset{¯}{!}\right)}_p$ is congruent (modulo $p^s$) to the product of thosenumbers $m$ where $m\equiv m-1(mod{p}^s)\leftrightarrow m^2\equiv 1(mod{p}^s)$.

Let $p$ be an odd prime and $s\in \mathbb{N}$. Since $2|̸p^s$, $p^s|(m^2-1)$ implies $p^s|(m+1)$ either or $p^s|(m-1)$. Thisleads to

$${\left(p^s\underset{¯}{!}\right)}_p\equiv -1(mod{p}^s)$$

for odd prime $p$ and any $s\in \mathbb{N}$.

Now let $p=2$ and $s\ge 2$. Then

$${(1+t{.2}^{s-1})}^2\equiv 1(mod{2}^s),t=\stackrel{-}{+}1.$$

Since

$$\left(2^{s-1}+1\right)\left(2^{s-1}-1\right)\equiv -1(mod{2}^s),$$

we have

$${\left(p^s\underset{¯}{!}\right)}_p\equiv (-1).(-1)=1(mod{p}^s)$$

For $p=2,s\ge 3$, but $-1$ for $s=1,2.\mathrm{\square }$