Prove that $\prod_{i=1}^st_i \equiv\ -1 \mod{P}$. Let $K$ be an algebraic number field and  $P$ be a prime ideal of $O_K$ and {$t_1,t_2,\ldots ,t_s$} be a system of representatives of all non-zero distinct elements of $O_K/P$. Prove that $\prod_{i=1}^st_i \equiv\ -1 \mod{P}$.
I have to prove that $1+\prod_{i=1}^st_i \in P$ but I cannot think of how, but if someone can supply a hint here, I ll be thankful.
 A: An strategy that works to prove Wilson's theorem works also in this general framework. 
If $p(x)$ is a polynomial in $O_K/P[x]$ of degree $n$ that vanishes at $n+1$ different points of $O_K/P$ then the coefficients of $p$ are all zero, in $O_K/P$. 
To see this we just write the $n+1$ equations. The determinant of the system (taking the coefficients of $p(x)$ as unknowns) is the Vandermonde determinant of the $n+1$ values. Since they are different and $P$ is prime, this determinant in invertible mod $P$. Therefore the coefficients must be zero. 
Now consider the polynomial $\prod_i(x-t_i)$ and $x^{s}-1$ they both vanish at $t_1,t_2,...,t_s$ and are monic. Therefore $\prod_i(x-t_i)-(x^{s}-1)$ has degree $s-1$ and vanishes at all $s$ points $t_1,...,t_s$. Therefore its coefficients are zero mod $P$.
In particular its constant coefficient $\prod_i t_i +1$ is zero mod $P$.
A: Recall that $\mathcal{O}_K/P$ is a finite field, and its multiplicative  group is thus cyclic.
Note that the cases that the characteristic is odd and that it is $2$ are somewhat different.
