$\omega$ is a n-th root of unity, aka $\omega^n=1$, calculate $$ D=\left| \begin{matrix} 1& \omega ^{-1}& \omega ^{-2}& \cdots& \omega ^{-n+1}\\ \omega ^{-n+1}& 1& \omega ^{-1}& \cdots& \omega ^{-n+2}\\ \omega ^{-n+2}& \omega ^{-n+1}& 1& \cdots& \omega ^{-n+3}\\ \vdots& \vdots& \vdots& & \vdots\\ \omega ^{-1}& \omega ^{-2}& \omega ^{-3}& \cdots& 1\\ \end{matrix} \right|. $$
My work: let $\omega_1,\cdots,\omega_n$ are roots of $x^n=1$, $f\left( x \right) =1+\omega ^{-1}x+\cdots +\omega ^{-n+1}x^{n-1}$, $$ V=\left| \begin{matrix} 1& 1& 1& \cdots& 1\\ \omega _1& \omega _2& \omega _3& \cdots& \omega _n\\ \omega _{1}^{2}& \omega _{2}^{2}& \omega _{3}^{2}& \cdots& \omega _{3}^{2}\\ \vdots& \vdots& \vdots& & \vdots\\ \omega _{1}^{n-1}& \omega _{2}^{n-1}& \omega _{3}^{n-1}& \cdots& \omega _{n}^{n-1}\\ \end{matrix} \right|, $$ then $$ DV=\left| \begin{matrix} f\left( \omega _1 \right)& f\left( \omega _2 \right)& f\left( \omega _3 \right)& \cdots& f\left( \omega _n \right)\\ \omega _1f\left( \omega _1 \right)& \omega _2f\left( \omega _2 \right)& \omega _3f\left( \omega _3 \right)& \cdots& \omega _nf\left( \omega _n \right)\\ \omega _{1}^{2}f\left( \omega _1 \right)& \omega _{1}^{2}f\left( \omega _2 \right)& \omega _{3}^{2}f\left( \omega _3 \right)& \cdots& \omega _{n}^{2}f\left( \omega _n \right)\\ \vdots& \vdots& \vdots& & \vdots\\ \omega _{1}^{n-1}f\left( \omega _1 \right)& \omega _{2}^{n-1}f\left( \omega _2 \right)& \omega _{3}^{n-1}f\left( \omega _3 \right)& \cdots& \omega _{n}^{n-1}f\left( \omega _n \right)\\ \end{matrix} \right|=f\left( \omega _1 \right) f\left( \omega _2 \right) \cdots f\left( \omega _n \right) V. $$ thus $D=f\left( \omega _1 \right) \cdots f\left( \omega _n \right) $.
My problem: $\omega$ is one of $\omega_1,\cdots,\omega_n$, how can I simplify the result $$\boxed{D=f\left( \omega _1 \right) \cdots f\left( \omega _n \right) }?$$
