Modular arithmetic. $(-1)^rr! \equiv 1 \pmod p$, then $(p-r-1)! \equiv -1 \pmod p$

Question

Let $p$ be a prime number and $r$ an integer such that $1 \le r \lt p$. If $(-1)^rr! \equiv 1 \pmod p$, then $(p-r-1)! \equiv -1 \pmod p$

I know that $n$ is a prime if and only if $(n -2)! \equiv 1 \pmod n$ and Wilson’s theorem: $p$ is a prime if and only if $(p - 1)! \equiv -1 \pmod p$. Still I haven’t found the right relation between them.

I would appreciate some help. Thanks.

Answer 1

Hint: $$\begin{align}(p-1)! &= (p-r-1)!\color{blue}{(p-r)(p-r+1)\cdots(p-1)}\\ &\equiv (p-r-1)!\color{blue}{(0-r)(0-r+1)\cdots(0-1)}\\ &=(p-r-1)!\color{blue}{(-r)(-(r-1))\cdots(-1)}\\ &=(p-r-1)!\color{blue}{(-1)^rr!}\pmod{p} \end{align}$$