Wilson's Theorem states:
$$ p \text{ is prime} \implies (p - 1)! ≡ -1 \mod p $$
The reduced residue set of prime \(p\) is \(\{ 1, 2, 3, \ldots, p-2, p-1 \}\). If we exclude \(1\) and \(p-1\), we get \(\{ 2, 3, \ldots, p-2 \}\). Each integer in this set has an inverse that is also in the set.
Now let \(R_p = \{ 1, 2, 3, \ldots, p-3 \}\). This means there is only one integer in \(R_p\) that doesn't have an inverse that is also in \(R_p\), Let's call this "inverseless" integer \(a\). The inverse of \(a\) is \(p-2\) but that doesn't exist in \(R_p\). Since every integer in \((p-3)!\) has an inverse except for \(a\):
$$\begin{gather} \frac{(p-3)!}{a} ≡ 1 \bmod p \\ (p-3)! ≡ a \bmod p \end{gather}$$
Now let's focus on \(a\). It is the only integer in \(R_p\) that doesn't have an inverse, because it's inverse is actually \(p-2\):
$$\begin{gather} a (p-2) ≡ 1 \bmod p \implies ap - 2a ≡ 1 \bmod p \\ ap ≡ 1 +2a \bmod p \end{gather}$$
Since \(ap\) is a multiple of \(p\):
$$\begin{gather} ap ≡ 0 \bmod p \implies 1 +2a ≡ 0 \bmod p \\ 2a ≡ -1 \bmod p \end{gather}$$
This means:
$$\begin{gather} (p-3)! ≡ a \bmod p \implies 2(p-3)! ≡ 2a \bmod p \\ 2(p-3)! ≡ -1 \bmod p \end{gather}$$