# Understanding of Proof of Wilson's Theorem

In my algebraic structures textbook there is a proof for the theorem if $p$ is a prime then $(p-1)!\equiv -1\pmod p\$.

Proof: Since p is prime, each element $1,2,3...(p-1)$ in $\Bbb Z_p$ has an inverse, hence the pairs of inverses will cancel out leaving only the self-inverse elements 1 and -1 and thus $(p-1)! = 1(-1) = -1$ in $\Bbb Z_p$.

My question is how is that possible when the elements of $(p-1)!$ are all positive? I must be confusing something...

• Do you know what the symbol $-1$ means in the context of $\mathbb{Z}_p$? It's something about divisibility, not negativity. – user296602 Jan 26 '16 at 2:54

"Positive" and "negative" don't make sense as comparisons to zero in $\mathbb{Z}_p$ - after all, if you add $1$ to itself $p$ times you get zero. And if you do it $p - 1$ times, you get something congruent to $-1$ modulo $p$. The statement that $(p - 1)! \equiv -1 \pmod p$ means that
$$p \mid (p - 1)! + 1$$
as a statement about divisors. What the proof is doing is collecting a bunch of pairs that multiply to $1$ modulo $p$ - that is, if you divide the product by $p$ you get a remainder of $1$.
$$(p-1)!\equiv -1\pmod p\$$ is the same as $$(p-1)!\equiv p-1\pmod p\$$ In $\pmod p$ negative doesn't mean what it does in your regular everyday arithmetic.