TITLE: Product of non-zero elements of a field is $-1$
QUESTION [0 upvotes]: This looks very much like something which is a duplicate, but I couldn't find the "desired duplicate". Sorry if this has been asked before.

Let $F$ be a field of $q$ elements and $F\setminus \{0\} = \{a_1,\ldots,a_{q-1}\}$. Show that $\prod a_i = -1$.

If $F$ has characteristic $2$ then this is trivial. So let's assume it has characteristic $p>2$. I tried to use the fact that $F\setminus \{0\}$ is a cyclic group with no avail. 
Any hint?

REPLY [4 votes]: Since you are working with a finite field with $q$ elements, anyone of them is a root of the following polynomial
$$ x^q-x=0.$$
In particular if we rule out the 0 element, any $a_i\neq 0$ is a root of 
$$x^{q-1}-1=0.$$
This polynomial splits completely in $\Bbb F_q$ so we find 
$$(x-a_1)\cdots (x-a_{q-1})=0$$
in particular $$x^{q-1}-1=(x-a_1)\cdots (x-a_{q-1})$$
Thus $a_1\cdots a_{q-1}=-1$.
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