Show that $2^r r! \equiv \prod_{k=1}^{r} [(-1)^k k)] \pmod{p}$

Question

Let be $\mathbb{P}$ the set of prime numbers, $p \in \mathbb{P} \cap (2\mathbb{Z} + 1)$, $r = \dfrac{p - 1}{2}$.

How do we proceed to show that $2^r r! \equiv \prod_{k=1}^{r} [(-1)^k k] \pmod{p}$.

My attempts started with:

\begin{align*} r! & = \prod_{k=0}^{r - 1} r - k \\ & = \prod_{k=0}^{r - 1} \dfrac{p - (2k + 1)}{2} \\ & = \dfrac{1}{2^r} \prod_{k=0}^{r - 1} p - (2k + 1) \end{align*}

Then I have shown that the following mapping is a bijection:

\begin{array}[lcl] & f & : & [[1, r]] \cap (2\mathbb{Z} + 1) & \longrightarrow & [[r + 1, p - 1]] \cap (2\mathbb{Z}) \\ & & x & \longmapsto & p - k \end{array}

I tried to rewrite $r!$ using $f$, but I couldn't do anything with it.

EDIT: Without supplementary laws of Gauss as I am currently trying to show them using this.

Answer 1

Note that $$2^r r!\equiv \prod_{k=1}^{r} (-1)^k k \equiv r!(-1)^{\sum\limits_{k=1}^{r} k}\pmod{p}$$

Now, as $r!$ is coprime to $p $, our problem becomes equivalent to $$2^r \equiv (-1)^{\frac{p^2-1}{8}} \pmod p$$

Now, as $\left(\frac{2}{p}\right) \equiv 2^r \pmod p$ by Euler's criterion, our next step is to show $$\left(\frac{2}{p}\right) \equiv (-1)^{\frac{p^2-1}{8}} \pmod p$$ Which is shown here.

Our proof is done.