設\(\displaystyle \omega=cos \frac{2\pi}{101}+i sin\frac{2\pi}{101}\)，求\((\omega-1)(\omega^2-1)\ldots (\omega^{100}-1)\)的值。
ans：101
請問要如何解出？謝謝！

\({{x}^{101}}=1\)的101個根為\(1,\omega ,{{\omega }^{2}},{{\omega }^{3}},\cdots ,{{\omega }^{100}}\)
\(\begin{align}
  & {{x}^{101}}-1=(x-1)(x-\omega )(x-{{\omega }^{2}})\cdots (x-{{\omega }^{100}}) \\
& {{x}^{100}}+{{x}^{99}}+\cdots +x+1=(x-\omega )(x-{{\omega }^{2}})\cdots (x-{{\omega }^{100}}) \\
& 101=(1-\omega )(1-{{\omega }^{2}})\cdots (1-{{\omega }^{100}})=(\omega -1)({{\omega }^{2}}-1)\cdots ({{\omega }^{100}}-1) \\
\end{align}\)