I have to prove that the product of all invertible elements in a finite field,$F^*$, equals $-1$. Now I know that $F^*$ is cyclic, so taking the product over $1,l,l^2,\ldots,l^{|F^*|}-1$, where $l$ generates $F^*$ really is the product of all invertible elements. But evaluating this product is a bit problematic if $|F^*|$ is even, because, although I know by little Fermat that $l^{|F^*|}=1$, I have to find out how many elements $a\in F^*$ there are, such that $a^2=1$ ; because for all other elements I can form pairs of different element $(l^e,l^f)$, that are inverse to each other and throw them in the product and that would yield 1. But I can't throw $a$ in the product, because on the other side of the $=$ I would have $-a$. If there is only one such $a$, namely $-1$, I would be finished.
After all this introduction, my question: How to prove, there is only one such $a$, namely $a=-1$ ? (And this question also hopefully motives my cryptic title)