How to complete this proof for Wilson's theorem I want to prove the first implication of Wilson's theorem, namely: 

For a prime number $p \geq 5$: $$(p-1)! \equiv -1 \mod p$$

I already proved: 


*

*$\forall 1 \leq n \leq p-1: \exists1 \leq k \leq p-1: n\cdot k \equiv1 \mod p$

*$\forall k \in \mathbb Z: (k^2 \equiv 1 \mod p \implies k \equiv \pm1 \mod p)$


How can I proceed from here?
Maybe write $(p-1)!$ as a product and show that two terms "cancel each other mod $p$" (using 1.). And show that there are a odd number of terms of the form described in 2. The product would then be congruent to $-1 \mod p$, but I don't really know how to fill the steps. 
 A: Your idea is right, they " cancel each other " but not every terms, it remains those which are their proper "inverse". In $(p-1)!$ It only stays the number $x$ satisfying
$$
x^2=1 \Leftrightarrow \left(x-1\right)\left(x+1\right)=0
$$
which gives you $x=1$ or $x=p-1$.
Hence

$$
\left(p-1\right)! \equiv -1 \left[p\right]
$$

A: So if $n\neq \pm 1\bmod p$ you can find a number $m\neq n$ with $mn=1 \bmod p$
(if $m=n$ you have $n^2=1\bmod p$ which you have shown occurs only for $\pm 1$)
Let's call $m$ a multiplicative inverse of $n$. What you have not yet done in what you have stated is to show that the multiplicative inverse is unique (modulo $p$). This is a necessary and straightforward thing to show for the proof to be complete.
Once this is done you can pair $2$ with its (unique) inverse, all the way up to $p-2$, at each stage pairing the least number which is not already paired. There are $p-3$ numbers, and since $p$ is odd, $p-3$ is even. Since these paired products are all equal to $1$ so is their total product.
Then you can multiply at the end by the missing numbers $1$ and $p-1=-1\bmod p$ to get your result.
