In the 5th edition, they have got the signs right, at least thats what the scanned version of the ebook says.

EDIT
We have $p-k \equiv -k \pmod{p}$ and hence $ \prod_{k=1}^{\frac{p-1}{2}} (p-k) \equiv \prod_{k=1}^{\frac{p-1}{2}} (-k) \pmod{p}$
Note that $(p-1)! = \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (p-k) \right)$
Hence, $(p-1)! \equiv \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (p-k) \right) \pmod{p} \equiv \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (-k) \right) \pmod{p}$
Hence, $(p-1)! \equiv (-1)^{\frac{p-1}{2}} \left(\prod_{k=1}^{\frac{p-1}{2}} k \right)^2 \pmod{p}$
$\left(\prod_{k=1}^{\frac{p-1}{2}} k \right) = \left(\frac{p-1}{2}\right)!$
From Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}$
Hence, $-1 \equiv (-1)^{\frac{p-1}{2}} \left[ \left(\frac{p-1}{2}\right)! \right]^2 \pmod{p}$
Hence, $\left[ \left(\frac{p-1}{2}\right)! \right]^2 \equiv (-1)^{\frac{p+1}{2}} \pmod{p}$
Hence, if $p = 4k+3$, then $(-1)^{\frac{p+1}{2}} = 1$
Hence, if $p = 4k+3$, $\left[ \left(\frac{p-1}{2}\right)! \right]^2 \equiv 1 \pmod{p}$