Proving Wilson's Theorem Using Gauss' Lemma

[mathsss2]

Use:
2\cdot4\cdot...\cdot(p-1)\equiv(2-p)(4-p)\cdot...\cdot(p-1-p)\equiv(-1)^{\frac{(p-1)}{2}}\cdot1\cdot3\cdot...\cdot(p-2) mod p
and
(p-1)!\equiv-1 mod p [Wilson's Theorem]
to prove
1^2\cdot3^2\cdot5^2\cdot...\cdot(p-2)^2\equiv(-1)^{\frac{(p-1)}{2}} mod p

Relevant equations

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

The attempt at a solution
need assistance

Thanks

 

[gabbagabbahey]

Hint:

2 \cdot 4 \cdot \ldots \cdot (p-1)=\frac{(p-1)!}{1 \cdot 3 \cdot \ldots \cdot (p-2)}=(p-1)! \cdot \left( \frac{1}{1 \cdot 3 \cdot \ldots \cdot (p-2)} \right)

 

[mathsss2]

Thanks, this problem is solved.