Proof of Wilsons theorem How to understand the grouping of integers from $2$ to $p-2$ in the proof af Wilsons theorem?

I don't understand how you can group the integers $2$ to $p−2$ into $(p-3)/2$ and next how you become the $2*3*...=1 (mod p)$

I think the best way to illustrate what's happening is to apply the proof to a specific value of $p$.  Take, for example, $p = 11$.
We can solve $2x \equiv 1 \pmod {11}$ with $x = 6$.  We can solve $3x \equiv 1 \pmod {11}$ with $x = 4$.  We can solve $5x = 1 \pmod {11}$ with $x \equiv 9$.  We can solve $7x \equiv 1 \pmod{11}$ with $x = 8$.
Note that in the above, we have accounted for all numbers from $2$ to $(11-2) = 9$.  We can now rearrange the product
$$
2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 =\\
(2 \cdot 6) \cdot (3 \cdot 4) \cdot (5 \cdot 9) \cdot (7 \cdot 8) \equiv\\
1 \cdot 1 \cdot 1 \cdot 1 \equiv 1 \pmod{11}
$$
That is, by grouping the integers from $2$ to $11 - 2$ into $\frac{11 - 3}{2} = 4$ pairs of multiplicative inverses, we have shown that 
$$
2 \cdot 3 \cdots (11 - 2)  \equiv 1 \pmod{11}
$$
which allows us to proceed with the proof.
