Wilson's Theorem states:
$$ p \text{ is prime} \implies (p - 1)! ≡ -1 \bmod p $$
It's obvious if \(p=2\), so we are going to assume that \(p\) is odd. We start by expanding \((p-1)!\):
$$ (1)(2)(3)(4) \ldots (p-3)(p-2)(p-1) $$
We know that \(1 ≡ 1 \mod p\) and \((p-1) ≡ -1 \mod p\), so we will ignore the first and last term for now and work with \((p-2)!\):
$$ (2)(3)(4) \ldots (p-4)(p-3)(p-2) $$
Every term in \(\{ 2,3,4, \ldots , (p-3), (p-2) \}\) is coprime with \(p\), and since \(p\) is odd, there is an even number of elements in \(\{ 2,3, \ldots , (p-2) \}\). There is no number in \(\{ 2,3, \ldots , (p-2) \}\) which is the inverse to \(1\) or \(p\), which means each number in the set has an inverse that is also in the set, so we can pair each number with their inverse:
$$ \begin{align} (2)(3) \cdots (p-3)(p-2) &= (a_1a_1^{-1})(a_2a_2^{-1}) \cdots (a_ka_k^{-1}) \\ &≡ (1)(1) \cdots (1) \mod p \end{align}$$
This means:
$$ \begin{align} (1)(2)(3)(4) \ldots (p-3)(p-2)(p-1) &≡ (1)(1)\ldots(1)(-1) \mod p \\ &≡ -1 \mod p \end{align}$$
Let's see an example with \(p=11\):
$$ 10! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10) ≡ ? \mod p $$
Let's pair the inverses:
$$ \begin{align} 10! = (1 \cdot 10)(2 \cdot 6)(3 \cdot 4) (5 \cdot 9)(7 \cdot 8) ≡ (-1)(1)(1)(1)(1) \mod p \end{align} $$