(restored slightly simplified version of earlier post). How about: $1- \zeta^{p-1} = -\zeta^{-1}(1-\zeta)$. Doing likewise for $\zeta^{2},\ldots,\zeta^{p-1}$, \zeta^{2},\ldots,\zeta^{\frac{p-1}{2}}$, and setting$\alpha = \prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i}),$we see that$\bar{\alpha} = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}} \alpha.$Hence we have $$p = \prod_{i=1}^{p-1}(1- \zeta^i) = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}}\prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i})^{2}.$$ 6 restored simplified version of earlier post Post deleted (restored slightly simplified version of earlier post). How about:$1- \zeta^{p-1} = -now redundant\zeta^{-1}(1-\zeta)$. Doing likewise for$\zeta^{2},\ldots,\zeta^{p-1}$, and setting$\alpha = \prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i}),$we see that$\bar{\alpha} = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}} \alpha.$Hence we have $$p = \prod_{i=1}^{p-1}(1- \zeta^i) = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}}\prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i})^{2}.$$ 5 removed post as redundant How about$\frac{1- \zeta^{p-1}}{1-\zeta} = (1 + \zeta + \ldots + \zeta^{p-2}) = -\zeta^{p-1}.$The same applies to$\zeta^2,\ldots,\zeta^{\frac{p-1}{2}}.\$ Hence we obtain $$p = \prod_{i=1}^{p-1}(1 - \zeta^i) = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2 -1}{8}} \prod_{i=1}^{\frac{p-1}{2}} (1  Post deleted- \zeta^{i})^{2}.$$now redundant