, 2016

Solution:
Define an $n \times n$ matrix $A_{n}(x)$ with entries $a_{i, j}=x$ if $i \equiv j \pm 1(\bmod n)$ and 1 otherwise. Let $F(x)=\sum_{\pi \in S_{n}}(-1)^{f(\pi)} x^{g(\pi)}$ (here $(-1)^{f(\pi)}$ gives the sign $\prod \frac{\pi(u)-\pi(v)}{u-v}$ of the permutation $\pi$ ). Note by construction that $F(x)=\operatorname{det}\left(A_{n}(x)\right)$.
We find that the eigenvalues of $A_{n}(x)$ are $2 x+n-2$ (eigenvector of all ones) and $(x-1)\left(\omega_{j}+\omega_{j}^{-1}\right)$, where $\omega_{j}=e^{\frac{2 \pi j i}{n}}$, for $1 \leq j \leq n-1$. Since the determinant is the product of the eigenvalues,
$$F(x)=(2 x+n-2) 2^{n-1}(x-1)^{n-1} \prod_{k=1}^{n-1} \cos \left(\frac{2 \pi k}{n}\right) .$$
Evaluate the product and plug in $x=-1$ to finish. (As an aside, this approach also tells us that the sum is 0 whenever $n$ is a multiple of 4.)