Converse of Wilson's theorem

Statement

Let $n$ be an integer with $n > 1$. If $(n - 1)! \equiv -1 \pmod n$, then $n$ is a prime number.

Remarks

With Wilson's theorem, this gives: an integer $n > 1$ is prime if and only if $(n - 1)! \equiv -1 \pmod n$. The proof: if $n = a b$ with $1 < a < n$, then $a$ divides $(n - 1)!$. A congruence $(n - 1)! \equiv -1 \pmod n$ would give $a \mid 1$, which is false. In fact $(n - 1)! \equiv 0 \pmod n$ for each composite $n > 4$, and $3! \equiv 2 \pmod 4$. This primality test is not practical, because $(n - 1)!$ needs about $n$ multiplications.

Explanations

L1 human

For a composite number $n$, a factor of $n$ appears in the product $1 \cdot 2 \cdots (n - 1)$. So the product cannot leave the remainder $n - 1$. For example, $5! = 120 \equiv 0 \pmod 6$.