How do you prove set with modulo? Given any prime $p$. Prove that $(p-1)! \equiv -1 \pmod p$.
How to prove this?

OK. The first thing you need to know is that if $x^2 \equiv 1 \pmod{p}$ where $p$ is a prime number then you have $x \equiv 1 \pmod{p}$ or $x\equiv -1 \pmod{p}$.
I'm going to state it as a Lemma and prove it:
Theorem: If $p$ is a prime number the only solutions to the congruence $x^2 \equiv 1 \pmod{p}$ are $x \equiv 1 \pmod{p}$ and $x \equiv -1 \pmod{p}$
Proof: 
$$x^2 \equiv 1 \pmod{p} \implies p \mid x^2-1 \implies p \mid (x-1)(x+1) \implies p \mid x-1 \text{ or } p \mid x+1 \implies x \equiv 1 \pmod{p} \text{ or } x \equiv -1 \pmod{p}$$
The last implication is correct because $p$ is a prime number.
This Lemma tells us that the only elements in a congruence modulo $p$ that are self-invertible are $1$ and $p-1 \equiv -1 \pmod{p}$.
Now, suppose that you multiply all non-zero elements $\pmod {p}$ as follows:
$A \equiv 1 \cdot 2 \cdot3 \cdots (p-2)(p-1) \pmod{p}$
Since $p$ is a prime number, all of these are non-zero and are relatively prime to $p$. So, they have inverses modulo $p$. Now we're going to use the lemma we just proved.
So, the case $p=2$ is obviously correct because $(2-1)!=1 \equiv -1 \pmod{2}$, so, you can assume that $p$ is an odd prime number. Since $p$ is odd, the number of the terms in $N=\{ 2, 3, \cdots, p-2 \}$ is even.
Now, just match each number in the set $N$ with its inverse modulo $p$. They all get canceled in pairs since there is an even number of them and none of them are self-invertible. At the end everything in $N$ gets canceled and becomes $1$ and you're left with only $1$ and $p-1$ and their product is obviously equal to $p-1$ which is the same as $-1$ modulo $p$. Just like the below:
$$1 \cdot \underbrace{ (2 \cdot 3 \cdots (p-3) \cdot (p-2))}_\text{Numbers in here get canceled in pairs} \cdot (p-1)  \equiv 1 \cdot (p-1) \equiv -1 \pmod{p}$$
