Wilson theorem Question Explanation

[mathsss2]

How do I explain this:

Let $$p$$ be odd prime explain why: $$2*4*...*(p-1)\equiv (2-p)(4-p)*...*(p-1-p)\equiv(-1)^{\frac{(p-1)}{2}}*1*3*...*(p-2)$$ mod $$p$$.



Relevant equations

Gauss lemma
wilson's theorem [$$(p-1)!\equiv-1$$ mod$$p$$]



The attempt at a solution
pairing? need assistance


Thanks

 

[gabbagabbahey]

Hint: What are $2-(2-p)$, $4-(4-p)$, ... $(p-1)-(p-1-p)$? Are they divisible by $p$?

 

[mathsss2]

Thanks, this problem is solved.