More Questions from Mathematical Analysis by Apostol

With the 2nd question, 2nd part, you are asked to

$$z^n - 1 = \prod\limits_{k = 1}^{n} \left( z - e^{\dfrac{2ki\pi}{n}} \right) \Rightarrow \prod\limits_{k = 1}^{n - 1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n - 1}}$$

Note that when $k=n$ $$e^{\frac{2ki\pi}{n}}=e^{2i\pi}=1$$ also $$z^n-1=(z-1)(z^{n-1}+z^{n-2}+z^{n-3}+...+z^2+z+1)$$ altogether $$\color{red}{(z-1)}(z^{n-1}+z^{n-2}+z^{n-3}+...+z^2+z+1)=\color{red}{(z-1)}\prod\limits_{k = 1}^{n-1} \left( z - e^{\frac{2ki\pi}{n}} \right)$$ which is $$z^{n-1}+z^{n-2}+z^{n-3}+...+z^2+z+1=\prod\limits_{k = 1}^{n-1} \left( z - e^{\frac{2ki\pi}{n}} \right)$$ and substituting $z=1$ $$n=\prod\limits_{k = 1}^{n-1} \left( 1 - e^{\frac{2ki\pi}{n}} \right)= \prod\limits_{k = 1}^{n-1} e^{\frac{ki\pi}{n}} \left( e^{-\frac{ki\pi}{n}} - e^{\frac{ki\pi}{n}} \right)=\\ (2i)^{n-1} (-1)^{n-1} \cdot \prod\limits_{k = 1}^{n-1} e^{\frac{ki\pi}{n}} \left(\frac{ e^{\frac{ki\pi}{n}} - e^{-\frac{ki\pi}{n}}}{2i} \right)=\\ 2^{n-1} (-i)^{n-1} \cdot \prod\limits_{k = 1}^{n-1} e^{\frac{ki\pi}{n}} \sin{\left(\frac{k\pi}{n}\right)}=\\ 2^{n-1} (-i)^{n-1} e^{\sum\limits_{k=1}^{n-1}\frac{ki\pi}{n}} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}=\\ 2^{n-1} (-i)^{n-1} e^{\frac{i\pi}{n}\frac{n(n-1)}{2}} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}=\\ 2^{n-1} (-i)^{n-1} i^{n-1} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}=2^{n-1} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}$$