What is the Galois group of $(x^2-1)(x^2-2)...(x^2-p+1)$ over $\mathbb Z\_p$ for an odd prime, $p$? There are exactly $\frac{p-1}{2}$ squares in $\mathbb Z\_p$ but my guess is the group is $\mathbb Z\_2$.
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By the uniqueness of finite fields of a given degree, there is one and **only** one field of order $p^2$. Since not all elements of $\Bbb F\_p$ are squares, you know that if $a$ is a non-square then

> 
> $$\Bbb F\_p[\sqrt{a}]=\Bbb F\_p[x]/(x^2-a)=\Bbb F\_{p^2}$$
> 
> 
> 

is an extension of degree $2$. But then since another non-square $b$ would also generate an extension $\Bbb F\_p[b]$ of degree $2$, it must be that these two are **the same field**, i.e. $\Bbb F\_p[\sqrt{b}]=\Bbb F\_{p^2}$ as well. So the splitting field for that polynomial is $\Bbb F\_{p^2}$ since that field has all the square roots of all elements of $\Bbb F\_p$ in it. The Galois group of this extension has order $2$, hence must be $\Bbb Z/2\Bbb Z$, as you surmise.
Also one can use the fact that $\mathbb{F}\_{p^2}/\mathbb{F}\_p$ is Galois, in other words is normal, so one can write $\prod (x^2-i)$ actually splits in $\mathbb{F}\_{p^2}$. So, ans is $\mathbb{Z}\_2$