Theorem: Factors of Sum of Two Even Powers

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$x^{2 n} + y^{2 n} = \ds \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {\paren {2 k - 1} \pi} {2 n} + y^2}$

Proof:
From Factorisation of $z^n + a$:
:$z^{2 n} + y^{2 n} = \ds \prod_{k \mathop = 0}^{2 n - 1} \paren {x - \alpha_k y}$
where $\alpha_k$ are the complex $2n$th roots of negative unity:
{{begin-eqn}}
{{eqn | l = \alpha_k
      | r = e^{i \paren {2 k + 1} \pi / {2 n} }
      | c = from Roots of Complex Number
}}
{{eqn | r = \map \cos {\dfrac {\paren {2 k + 1} \pi} {2 n} } + i \, \map \sin {\dfrac {\paren {2 k + 1} \pi} {2 n} }
      | c = $k \in \set {0, 1, 2, \ldots, 2 n - 1}$
}}
{{end-eqn}}
Then we have that:
:$U_{2 n} = \set {\tuple {\alpha_0, \alpha_{2 n - 1} }, \tuple {\alpha_1, \alpha_{2 n - 2} }, \ldots, \tuple {\alpha_k, \alpha_{2 n - k - 1} }, \ldots, \tuple {\alpha_{n - 1}, \alpha_n } }$
where $U_{2 n}$ denotes the complex $2n$th roots of negative unity:
:$U_{2 n} = \set {z \in \C: z^{2 n} = -1}$
Taking the product, $p_k$, of the factors of $x^{2 n} + y^{2 n}$ in pairs:
{{begin-eqn}}
{{eqn | l = p_k
      | r = \paren {x - \alpha_k y} \paren {x - \alpha_{2 n - k - 1} y}
      | c = 
}}
{{eqn | r = \paren {x - \alpha_k y} \paren {x - \overline {\alpha_k} y}
      | c = Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs
}}
{{eqn | r = x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + \alpha_k y \overline {\alpha_k} y
      | c = 
}}
{{eqn | r = x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + \cmod {\alpha_k}^2 y^2
      | c = Modulus in Terms of Conjugate
}}
{{eqn | r = x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + y^2
      | c = Modulus of Complex Root of Negative Unity equals 1
}}
{{eqn | r = x^2 - x y \paren {\cos \dfrac {\paren {2 k + 1} \pi} {2 n} + i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} + \cos \dfrac {\paren {2 k + 1} \pi} {2 n} - i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} } + y^2
      | c = Definition of $\alpha_k$
}}
{{eqn | r = x^2 - 2 x y \cos \dfrac {\paren {2 k + 1} \pi} {2 n} + y^2
      | c = simplification
}}
{{end-eqn}}
However
{{begin-eqn}}
{{eqn | l = p_{n - k}
      | r = x^2 - 2 x y \cos \dfrac {\paren {2 n - 2 k + 1} \pi} {2 n} + y^2
      | c =
}}
{{eqn | r = x^2 - 2 x y \cos \dfrac {\paren {-2 k + 1} \pi + 2 n \pi} {2 n} + y^2
      | c =
}}
{{eqn | r = x^2 + 2 x y \cos \dfrac {\paren {-2 k + 1} \pi} {2 n} + y^2
      | c = Cosine of Angle plus Straight Angle: $\map \cos {x + \pi} = -\cos x$
}}
{{eqn | r = x^2 + 2 x y \cos \dfrac {\paren {2 k - 1} \pi} {2 n} + y^2
      | c = Cosine Function is Even: $\map \cos {- x} = \cos x$
}}
{{end-eqn}}
Consider the permutation:
:<nowiki>$\sigma = \begin{pmatrix}
  1 & 2     & \cdots & k     & \cdots & n - 1 & n \\
  n & n - 1 & \cdots & n - k & \cdots & 2     & 1
\end{pmatrix}$</nowiki>
From Permutation of Indices of Product:
:$\ds \prod_{\map R k} p_k = \prod_{\map R {\map \sigma k} } p_{\map \sigma k}$
Hence:
{{begin-eqn}}
{{eqn | l = x^{2 n} + y^{2 n}
      | r = \prod_{k \mathop = 1}^n p_k
      | c =
}}
{{eqn | r = \prod_{k \mathop = 1}^n p_{n - k}
      | c =
}}
{{eqn | r = \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {\paren {2 k - 1} \pi} {2 n} + y^2}
      | c =
}}
{{end-eqn}}
{{Proofread|I believe this to be correct, but would welcome a sanity check.}}
{{qed}}