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My idea is to define a $(n-1)\times(n-1)$ Hermitian matrix with diagonal elements equal to zero and off-diagonal elements $a_{ij}= 1/(1-x_{i-j})$, where $x_k=\zeta^k$. So the left-hand side of (4) is equal to $\det(A)$. Multiply $A$ from right by a $(n-1)\times(n-1)$ diagonal matrix $b_{ii}=(1-x_{is})$ for any fixed $s\in\{-(n-1)/2,\ldots,-1,1,(n-1)/2\}$. We can show the resulting matrix $C_s$ has an eigenvalue $s$ with eigenvector whose $j$-th element is equal to $x_{j((n-1)/2-s)}$. The difficulty is to show such an eigenvalue of $C_s$ is also an eigenvalue of $C_1$ for all s, even though numerical evaluation has suggested it. Prof. Fedor Petrov has a very elegant proof of this. See here.