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

$$\alpha^{2n}-1=\prod_{k=0}^{k=2n-1}(\alpha-e^{i\frac{2k\pi}{2n}})$$ 
$$\prod_{k=0}^{k=2n-1}(\alpha-e^{i\frac{2k\pi}{2n}})=(\alpha-1)(\alpha+1)\prod_{k=1}^{k=n-1}(\alpha-e^{i\frac{k\pi}{n}})(\alpha-e^{-i\frac{k\pi}{n}})$$

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*}
 A: The first line is a statement about roots of unity and factorization. The numbers $e^{2\pi i k / n}$ are precisely the $n$th roots of unity. (And the factor theorem, I suppose).
The second one is just a rearrangement of terms from the first. If you'd like an alternate view, you might think of $\alpha$ as a variable, and consider each side as polynomials in $\alpha$. Then since they have the same roots and the same value at $\alpha = 0$, they are the same polynomial.
In short: you should read up on roots of unity and possibly the factor theorem. 
A: First Identity: Let $$p(x) = x^n + a_{n-1} x^{n-1}+ \ldots+a_0$$ a polynomial function of order $n$. Then, by the fundamental theorem of algebra, we have $$p(x) = \prod_{j=1}^n (x-x_j)$$ where $x_1,\ldots,x_n$ are the roots of $p$. Apply this to $p(x) := x^{2n}-1.$
Second Identity:
Write $$\prod_{k=0}^{2n-1} \left(\alpha - e^{\imath \frac{2k\pi}{2n}} \right) = \underbrace{\left(\prod_{k=0}^0 \dots \quad \cdot \prod_{k=n}^n \dots \right)}_{(\alpha-1)(\alpha+1)} \left( \prod_{k=1}^{n-1} \dots \quad \cdot \prod_{k=n+1}^{2n-1}  \dots \right).$$
Now note that
$$\exp \left( 2\pi \imath \frac{2n-\ell}{2n} \right) = \exp \left( 2\pi \imath - 2\pi \imath \frac{\ell}{2n} \right) = \exp \left(- 2\pi \imath \frac{\ell}{2n} \right) \tag{1}$$
as $x \mapsto e^{\imath x}$ is periodic. For $k \in \{n+1,\ldots,2n-1\}$ we write $k = 2n-\ell$ where $\ell \in \{1,\ldots,n-1\}$. Applying $(1)$, we get
$$\prod_{k=n+1}^{2n-1} \left(\alpha - e^{\imath \frac{2k\pi}{2n}} \right) = \prod_{\ell=1}^{n-1} \left(\alpha - e^{- \imath \pi \frac{\ell}{n}} \right).$$
This finishes the proof. 
