Wilson's Theorem

Theorem
A positive integer $$p$$ is a prime iff $$\left({p-1}\right)! \equiv -1 \pmod {p}$$.

Proof
If $$p = 2$$ the result is obvious.

Therefore we assume that $$p$$ is an odd prime.

If part
First let $$p$$ be a prime.

Since modular inverse modulo $$p$$ is a bijection in which only 1 and $$p-1$$ are mapped to themselves, numbers $$2, \ldots, p-2$$ can be divided into pairs $$\left({a, b}\right)$$, such that $$a b \equiv 1$$.

The product of all these numbers is therefore $$1$$.

By definition, $$\left({p-1}\right)!$$ is the product of $$1$$,$$p-1$$ and the numbers listed above.

Because of that, $$\left({p-1}\right)!$$ is congruent to $$1 \cdot \left({p-1}\right) \cdot 1 \equiv -1 \pmod {p}$$.

Only if part
Now consider $$p$$ is a composite, and $$q$$ is a prime such that $$q \backslash p$$.

Then both $$p$$ and $$\left({p-1}\right)!$$ are divisible by $$q$$.

If the congruence $$\left({p-1}\right)! \equiv -1 \pmod{p}$$ was satisfied, we would have $$\left({p-1}\right)! \equiv -1 \pmod {q}$$.

This would means $$0 \equiv -1 \pmod {q}$$ which is false.

Hence for $$p$$ composite, the congruence $$\left({p-1}\right)! \equiv -1 \pmod {p}$$ cannot hold.

Historical Note
This proof was attributed to John Wilson by Edward Waring in his 1770 edition of "Meditationes Algebraicae".

However, it appears to have been known to Gottfried Leibniz before 1663.