In Wilson's Theorem Why can we always split integers into mutually inverse pairs?

Question

I mean for example $p=13$

$1*12\equiv-1 \pmod{13}$

then inverse pairs

$2*7\equiv1\pmod{13}$

$3*9\equiv1\pmod{13}$

$4*10\equiv1\pmod{13}$

$5*8\equiv1\pmod{13}$

$6*11\equiv1\pmod{13}$

Answer 1

I assume $p$ is an odd prime. Then you have the residues 1,2,..,p-1. Now very $a \in \{1,.., p-1\}$ has an inverse $a^{-1}$ with $a a^{-1} \equiv 1 \pmod p.\; $But $a\equiv 1 \pmod p$ and $a\equiv -1\equiv p-1 \pmod p$ are self-inverse (and they are the only two roots of unity), and the remaining $p-3$ residues come in pairs with $a a^{-1} \equiv 1 \pmod p,\,$ and $a \not \equiv a^{-1} \pmod p$. Therefore $(p-1)! \pmod p$ is the product $$(p-1)! \equiv (1)(-1)\prod_{k=1}^{\frac{p-3}{2}}1 \equiv -1\pmod p$$

Note: For $p=13\;$ you have the roots of unity 1 and 12, as well as the pairs (2,7), (3,9), (4,10), (5,8), and (6,11).