Proving that:$x^{2n}-1=(x^2-1)\prod_{k=1}^{n-1}(x^2-2x\cos{\frac{k\pi}{n}}+1)$ Proving that: $$x^{2n}-1=(x^2-1)\prod_{k=1}^{n-1}(x^2-2x\cos{\frac{k\pi}{n}}+1)$$
I have the solution but I am having trouble understading it, or I think that it is not correct, because this is a notebook from class, so I guess there's always a possibility for error. I will type out the steps: All are unclear to me accept the fact that $\sqrt[2n]{1}=e^{\frac{2k\pi}{2n}i}.$
$$x^{2n}=1(?) \\\sqrt[2n]{1}=e^{\frac{2k\pi}{2n}i},n=0,1,...,2n-1  \\ x^{2n}-1=\prod_{k=1}^{2n-1}(x-e^{\frac{2k\pi}{2n}i})=(x+1)(x-1)\prod_{k=1}^{n-1}(x-e^{\frac{2k\pi}{2n}i})(x-e^{\frac{2(2n-k)\pi}{2n}i}) \ \ \ (?) \\ ... \\ ...$$
It is very important for me to understand this path in solving this problem, because these techniques are probably going to apear on the future exam.
 A: We have to factorize the polynomial $x^{2n}-1$. In $\mathbf C$ this is the same as finding its roots, so we have to solve $x^{2n}=1$. The solutions to this equation are $e(k/2n)$ ($k=0,\dots,2n-1$), where $e(t):=e^{i2t\pi}$, Therefore, we have the factorization
$$x^{2n}-1=\prod_{0\leq k\leq 2n-1}(x-e(k/2n)).$$
But $e(0)=1$ and $e(n/2n)=-1$, so
$$\prod_{0\leq k\leq 2n-1}(x-e(k/2n))=(x-1)(x+1)\prod_{1\leq k\leq 2n-1,k\neq n}(x-e(k/2n))=(x^2-1)\prod_{1\leq k\leq 2n-1,k\neq n}(x-e(k/2n)).$$
Now,
$$\begin{align}
\prod_{1\leq k\leq 2n-1,k\neq n}(x-e(k/2n))&=
\prod_{1\leq k\leq n-1,k\neq n}(x-e(k/2n))
\prod_{n+1\leq k\leq 2n-1,}(x-e(k/2n))\\
%&=\prod_{1\leq k\leq n-1}(x-e(k/2n))
%\prod_{1\leq k\leq n-1}(x-e((k+n)/2n))\\
&=\prod_{1\leq k\leq n-1}(x-e(k/2n))
\prod_{1\leq k\leq n-1}(x-e((2n-k)/2n))\tag{$*$}\label{eq1}\\
&=\prod_{1\leq k\leq n-1}(x-e(k/2n))
\prod_{1\leq k\leq n-1}(x-e(-k/2n))\\
&=\prod_{1\leq k\leq n-1}(x-e(k/2n))
(x-e(-k/2n))\\
&=\prod_{1\leq k\leq n-1}(x^2-x(e(k/2n)+e(-k/2n))+1)\\
\end{align}$$
But $e(x)+e(-x)=2\cos(2\pi x)$, so
$$\begin{align}
\prod_{1\leq k\leq n-1}(x^2-x(e(k/2n)+e(-k/n))+1)&=
\prod_{1\leq k\leq n-1}(x^2-2x\cos(2\pi k/2n)+1)\\
&=\prod_{1\leq k\leq n-1}(x^2-2x\cos(\pi k/n)+1).
\end{align}$$
Therefore
$$x^{2n}-1=
(x^2-1)
\prod_{1\leq k\leq n-1}(x^2-2x\cos(\pi k/n)+1).$$
EDIT (more steps in equation \eqref{eq1})
$$\begin{align}
\prod_{n+1\leq k\leq 2n-1,}(x-e(k/2n))&=
\left(x-e\left(\frac{n+1}{2n}\right)\right)
\left(x-e\left(\frac{n+2}{2n}\right)\right)
\dots
\left(x-e\left(\frac{n+(n-1)}{2n}\right)\right)\\
&=
\left(x-e\left(\frac{n+(n-1)}{2n}\right)\right)
\dots
\left(x-e\left(\frac{n+2}{2n}\right)\right)
\left(x-e\left(\frac{n+1}{2n}\right)\right)\\
&=
\left(x-e\left(\frac{2n-1}{2n}\right)\right)
\dots
\left(x-e\left(\frac{2n-(n-2)}{2n}\right)\right)
\left(x-e\left(\frac{2n-(n-1)}{2n}\right)\right)\\
&=\prod_{1\leq k\leq n-1}\left(x-e\left(\frac{2n-k}{2n}\right)\right)
\end{align}$$
A: Using $$z^n-1=(z-1)(z-\alpha)(z-{\alpha}_2)\cdots(z-{\alpha}_{n-1})$$
$$(x-\alpha)(x-\alpha')=x^2-2x\cos\frac{\pi}{n}+1$$
