I need to use the following identities for poisson integral but i can't guz i don't know how to prove them.

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> $$\alpha^{2n}-1=\prod\_{k=0}^{k=2n-1}(\alpha-e^{i\frac{2k\pi}{2n}})$$ 
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> $$\prod\_{k=0}^{k=2n-1}(\alpha-e^{i\frac{2k\pi}{2n}})=(\alpha-1)(\alpha+1)\prod\_{k=1}^{k=n-1}(\alpha-e^{i\frac{k\pi}{n}})(\alpha-e^{-i\frac{k\pi}{n}})$$
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I would appreciate any help

what i did :

Let $\alpha >1\,\quad \alpha^{2n} =1$

\begin{align\*}
\alpha^{2n} &=1\\
(\rho.e^{i\phi})^{2n} &=1.e^{i.0} \text{ with $\rho \in \mathbb{R}^{+}\_{\*},\ \phi \in \mathbb{R}$}\\
\rho^{2n}.e^{i.2n.\phi}&=1.e^{i.0}\\
\rho^{2n}&=1, \quad e^{i.2n\phi}=e^{i.0}\\
\rho&=1, \quad 2n.\phi=0[2\pi]\\
 \rho&=1, \quad \exists k\in \mathbb{Z} \ 2n.\phi=2.k.\pi\\
 \rho&=1, \quad \phi=\frac{2.k.\pi}{2n} \text{ Then } \alpha\_{k}=e^{i\frac{2.k.\pi}{2n}}\\
\text{Let } k &= t + 2nq \text{ or } t\in \{0,1,\ldots,2n-1\} \text{ and } q \in \mathbb{Z}.\\
&\text{ Then } \alpha\_{k}=e^{i\dfrac{2.t.\pi}{2n}+2.\pi.q}=Z\_{t} 
\\
S&=\{Z\_{k}=e^{i\dfrac{2.k.\pi}{2n}}|k\in \{0,1,\ldots,2n-1\} \}
\end{align\*}