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!

$$1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p-1}{2}} \pmod p$$
is the same as
$$1^2 2^2 3^2 \cdots
\left(\frac{p-1}{2}\right)^2
\equiv (-1)^{\frac{p-1}{2}} \pmod p,$$
since we are working modulo $p$.
Now we can split each square itself as $ i^2 = i(p-i)(-1) $
$$(-1)^{\frac{p-1}{2}} 123\cdot\cdot\cdot(p-1)\equiv (-1)^{\frac{p+1}{2}} \pmod p$$
from Wilson's theorem.
