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 $p = 4k+3$, $$\left[ \left(\frac{p-1}{2}\right)! \right]^2 \equiv 1 \pmod{p}$$