The root-of-unity derangement sum conjecture

Let $n>1$ be an integer, let $\zeta$ be a primitive $n$-th root of unity, and define the derangement set

$$D(n):=\{\tau\in S_n:\tau(j)\ne j\text{ for all }j=1,\ldots,n\}.$$

For a permutation $\tau$, let $\operatorname{sign}(\tau)$ denote its sign. The root-of-unity derangement sum conjecture.** If $n$ is even, then

$$\sum_{\tau\in D(n)}\prod_{j=1}^n\frac1{1-\zeta^{j-\tau(j)}}=\frac{((n-1)!!)^2}{2^n}=\frac{n!}{4^n}\binom n{n/2}.$$

If $n$ is odd, then

$$\sum_{\tau\in D(n-1)}\prod_{j=1}^{n-1}\frac1{1-\zeta^{j-\tau(j)}}=\frac1n\left(\frac{n-1}{2}!\right)^2,$$ $$\sum_{\tau\in D(n-1)}\operatorname{sign}(\tau)\prod_{j=1}^{n-1}\frac1{1-\zeta^{j-\tau(j)}}=\frac{(-1)^{(n-1)/2}}n\left(\frac{n-1}{2}!\right)^2,$$

and

$$\sum_{\tau\in D(n-1)}\operatorname{sign}(\tau)\prod_{j=1}^{n-1}\frac{1+\zeta^{j-\tau(j)}}{1-\zeta^{j-\tau(j)}}=\frac{(-1)^{(n-1)/2}}n((n-2)!!)^2.$$

The conjecture is motivated by a preceding theorem; the supplied text gives no resolution of these identities.