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 $((\frac{p-1}{2})!)^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$

Suppose $\;p=4k+1\;$ , so that
$$\frac{p-1}2=2k\implies\;\; \text{since}\;\;j=p-j\pmod p\;\;\text{for}\;\;0\le j\le p \,, $$
we get that
$$\frac{p-1}2+1=\frac{p-1}2-1\pmod p\;,\;\;\frac{p-1}2+2=\frac{p-1}2-2\pmod p\,,\ldots$$
so that
$$1\cdot2\cdot\ldots\cdot k\cdot\ldots\cdot\left(\frac{p-1}2-1\right)\cdot\left(\frac{p-1}2\right)\cdot\left(\frac{p-1}2-1\right)\cdot\ldots\cdot k\cdot\ldots \cdot1=\left[\left(\frac{p-1}2\right)!\right]^2$$
If you know some basic group theory things can possibly go easier than the above.
