Decomposition of polynomial of n-degree into irreducible polynomials. I'm getting ready for the exam in algebra. And I have problem that I can't solve again.
I'm understand what I need to do if I have a polynomial with a given degree. But I very confused by arbitrary degree in this problem. 

Given polynomial
  $$x^{2n}-2x^n+2$$
  The task is find expansion of the polynomial as a product of irreducible polynomials in $\mathbb{R}$.

Thanks in advance.

$$x^{2n}-2x^n+2=(x^n-1)^2+1=(x^n-1+i)(x^n-1-i)=(x^n-2^{0.5}e^{\frac{-i\pi}{4}})(x^n-2^{0.5}e^{\frac{i\pi}{4}})$$
Now note that:
$$x^n-2^{0.5}e^{\frac{-i\pi}{4}}=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})$$
$$x^n-2^{0.5}e^{\frac{i\pi}{4}}=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}})$$
Hence:
$$x^{2n}-2x^n+2=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}})(x-2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})$$
$$x^{2n}-2x^n+2=\prod_{k=1}^n(x^2-(2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}}+2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})x+2^{\frac{1}{n}})$$
$$x^{2n}-2x^n+2=\prod_{k=1}^n(x^2-2(2^{\frac{1}{2n}}\cos({\frac{\pi}{4}+\frac{2\pi k}{n}}))x+2^{\frac{1}{n}})$$
