Htykuut

$1^2 cdot 3^2 cdot 5^2cdots (p-2)^2equiv (-1)^{frac{p+1}{2}}pmod p$

I saw that I can use Wilson's theorem that $(p-1)!equiv -1 pmod p$ and that they change something, they said put on even number $frac{p+1}{2}$ so now you prove it but I do not understand can you help me?

We work in the field $Bbb F_p$ with $p$ elements. (Same as $Bbb Z/p$, for the given prime $p$, i suppose it is a prime.)

Then
$$
begin{aligned}
-1 &= 1cdot 2cdot 3cdot 4cdot 5cdot 6cdot dots cdot (p-2)cdot(p-1)
\
&=1color{red}{cdot (-2)}cdot 3color{red}{cdot (-4)}cdot 5color{red}{cdot (-6)}cdot dots cdot (p-2)color{red}{cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (-2)cdot (-4)cdot (-6)cdot dots cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (p-2)cdot (p-4)cdot (p-6)cdot dots cdot(p-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=(1cdot 3cdot 5cdot dots cdot (p-2))^2
color{blue}{cdot(-1)^{(p-1)/2}}
.
end{aligned}
$$
This leads to the answer...