Show that: $-1 \equiv (-1)^{(p-1)/2}x^2 \pmod p$ I'm currently working in the following Wilson's theorem excercise:

Being $p$ an odd prime. Let $x=((p-1)/2)!$
Using the fact that $(p − 1)! =
(1(p − 1)) (2(p − 2)) · · · (((p − 1)/2)((p + 1)/2))$ show that: $$-1 \equiv (-1)^{(p-1)/2}x^2 \pmod p$$

I'm starting from:
$$(1(p − 1)) (2(p − 2)) · · · (((p − 1)/2)((p + 1)/2))\equiv -1 \pmod p$$
My plan is to simplify terms on the left side to reach that $(p-1)/2$ which is the power of that $-1$, but I really don't know how can it be raised to that power, any hint or help will be really appreciated.
 A: As Lord Shark the Unknown implies in the question comment, note that in each set of brackets, you have factors, congruent to $p$, of $k$ and $-k$ going from $k = 1$ to $k = \left(p−1\right)/2$, e.g., in $\left(1\left(p - 1\right)\right)$, it is $1$ and $p - 1 \equiv -1 \pmod p$ and in $\left(2\left(p - 2\right)\right)$, it is $2$ and $p - 2 \equiv -2 \pmod p$, all the way to $\left(\left(\left(p−1\right)/2\right)\left(\left(p+1\right)/2\right)\right)$, it is $\left(p−1\right)/2$ and $\left(p + 1\right)/2 \equiv -p + \left(p + 1\right)/2 \equiv -\left(p - 1\right)/2 \pmod p$. As such, each factor of $1, 2, \ldots, \left(p−1\right)/2$ is repeated twice (so altogether the product is $x^2$), plus you have $\left(p−1\right)/2$ factors of $-1$. Using this along with Wilson's Theorem that
$$\left(p-1\right)! \equiv -1 \pmod p$$
gives that
$$\left(p-1\right)! \equiv -1 \equiv \left(-1\right)^{\left(p-1\right)/2}x^2 \pmod p$$
A: How about this:
\begin{align}
(p-1)! &= \prod_{k=1}^{(p-1)/2}k \cdot \prod_{k=1}^{(p-1)/2}(p-k)\\
& \equiv \prod_{k=1}^{(p-1)/2}k \cdot \prod_{k=1}^{(p-1)/2}(-k) = (-1)^{(p-1)/2}\cdot \prod_{k=1}^{(p-1)/2}k \cdot \prod_{k=1}^{(p-1)/2}k \pmod p\\
& \equiv (-1)^{(p-1)/2} \cdot x \cdot x = (-1)^{(p-1)/2}x^2 \pmod p
\end{align}
