Wilson's Theorem states:
$$ n \text{ is prime} \implies (n - 1)! ≡ -1 \bmod n $$
The converse of this would be:
$$ (n - 1)! ≡ -1 \bmod n \implies n \text{ is prime} $$
Let \(n = ab\) where \(a\) and \(b\) are integers, and let \(a \le \sqrt{n}\).
$$ 1 \le a \le \sqrt{n} \implies a|(n-1)! $$
Since \((n - 1)! ≡ -1 \bmod n\). This means:
$$ n | (n - 1)! + 1 $$
Since \(a|n\):
$$\begin{gather} a | (n - 1)! + 1 ∧ a|(n-1)! \\ \implies a|((n-1)! + 1) - (n-1)! \implies a|1\end{gather}$$
This means if \(a \le \sqrt{n}\), then \(a=1\), which means \(n\) is prime.