When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble - trouble understanding proof Theorem: When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble
Proof: When $p=2$, the statement is clear.
Assume $p\equiv 1\pmod{4}$, let $r=\frac{p-1}{2}$ and $x=r!$
Then since $r$ is even $x^2\equiv (1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
Thus, when $p\equiv 1\pmod{4}$, the congruence $x^2\equiv -1\pmod{p}$ is soluble.
Point of contention: I understand the general argument
I understand the relation
$(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
But I cant work out how $x^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
How is $((\frac{p-1}{2})!)^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$

If the general case proves obfuscatory, it often proves enlightening to first examine a few small special cases in order to help grasp the general idea, e.g. let's examine the two smallest cases first $\,p=5,13.\,$  mod $\, p = 5\!:\,\ \color{#0a0}{4\equiv -1},\ 3\equiv \color{#c00}{ -2}\ $ so substituting these into Wilson's formula  
$\qquad\qquad \begin{align} {\rm mod}\ 5\!:\,\ {-}1\, \equiv& \ \ \ \ \color{#0a0}{(4)}\cdot \color{}{(3)}\cdot 2\cdot 1\\
\equiv&\, \color{#0a0}{(-1)}(-\color{#c00}2)\cdot\color{#c00} 2\cdot 1\,\equiv\, \color{#c00}2!^2\end{align}$  
$\qquad\quad\ \ \begin{align} {\rm mod}\ 13\!:\,\ {-}1\, \equiv& \ \  \color{#0a0}{(12)}\cdot\,(11)\,\cdots\ \ (7)\ \cdot 6\cdot 5\cdots 1\\
\equiv&\, \color{#0a0}{(-1)}\cdot(-2)\cdots (-\color{#c00}6)\cdot\color{#c00} 6\cdot 5\cdots 1\,\equiv\, \color{#c00}6!^2 \end{align}$
$\qquad\qquad\quad\overset{\vdots}{\phantom{c}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ \overset{\ddots}{\phantom{c}}$
$\qquad\begin{align} {\rm mod}\ 4k\!+\!1\!:\,\ {-}1\, \equiv& \ \  \color{#0a0}{(4k)}(4k\!-\!1)\cdots (2k\!+\!1)\ \  (2k)(2k\!-\!1)\cdots 1\\
\equiv&\, \underbrace{\color{#0a0}{(-1)}\,\cdot\,(-2)\, \cdots\  \,(-\color{#c00}{2k})}_{\Large (-1)^{2k}(2k)!\ =\ (2k)!} \ \ \,\underbrace{(\color{#c00}{2k})(2k\!-\!1)\cdots 1}_{\Large(2k)!}\equiv (2k)!^2 \end{align}$
