Let $p$ be an odd prime, and $\zeta$ a primitive $p$-th root of unity over a field of characteristic $0$.
Let $G = \sum\limits_{j=0}^{p-1} \zeta^{j\left(j-1\right)/2}$ be the standard Gauss sum for $p$. (An alternative definition for $G$ is $G = \sum\limits_{j=1}^{p-1}\left(\frac{j}{p}\right)\zeta^j$, where the bracketed fraction denotes the Legendre symbol.) Denote by $k$ the element of $\left\lbrace 0,1,…,p-1\right\rbrace$ satisfying $16k\equiv -1\mod p$.
Then, $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right) = \zeta^k G$.
Question: Can we prove this identity purely algebraically, with no recourse to geometry and analysis?
If we can, then we obtain an easy algebraic proof for the value – including the sign – of the Gauss sum $G$, since both the modulus and the argument of $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right)$ are easy to find (mainly the argument – it's a matter of elementary geometry).
Note that my "algebraically" allows combinatorics, but I am somewhat skeptical in how far combinatorics alone can solve this. Of course, we can formulate the question so that it asks for the number of subsets of $\left\lbrace 1,2,…,\frac{p-1}{2}\right\rbrace$ whose sum has a particular residue $\mod p$, but whether this will bring us far… On the other hand, $q$-binomial identities might be of help, since $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right)$ is a $\zeta$-factorial.
I am pretty sure things like this must have been done some 100 years ago.
Best Answer
Have you seen this blog post of mine? Summary: Use Geoff's argument to prove this up to a sign; then use $p$-adic arguments to nail down the sign.