For prime $p>2: 1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$ 
Possible Duplicate:
Why is the square of all odds less than an odd prime $p$ congruent to $(-1)^{(p+1)/(2)}\pmod p$?
If p is an odd prime, prove that $1^2 \times 3^2 \times 5^2 \cdots  \times (p-2)^2 \equiv (-1)^{(p+1)/2}\pmod{p}$  

I'd love your help with proving the following claim:
For prime $p>2$:
$$1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p.$$
I instantly thought of Wilson Theorem which says that $1\cdot2\cdot3\cdot\cdot\cdot\cdot(p-1) \equiv (-1) \pmod p$, but I can't see how to use it.
I also tried to divide it to two cases, for $p \equiv 1 \pmod4$, and $p \equiv 3 \pmod4$, but again I didn't reach the conclusion.
Thanks a lot!

You know that 
$(p-1)! \equiv -1 \mod p$
Then, since $p-k\equiv -k \mod p$ we have
$$-1=(p-1)! =[1  \cdot 3 \cdot 5 \cdot ... \cdot (p-2)] \cdot [ 2 \cdot 4 ... \cdot (p-1)]$$
$$=[1  \cdot 3 \cdot 5 \cdot ... \cdot (p-2)] \cdot [ (p-2) \cdot (p-4) ... \cdot (1) \cdot (-1)^\frac{p-1}{2}] $$
