Factorisation of z^n+1

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:


 * $z^n + 1 = \displaystyle \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} n}$

Proof
From Factorisation of $z^n - a$, setting $a = -1$:


 * $z^n + 1 = \displaystyle \prod_{k \mathop = 0}^{n - 1} \paren {z - \alpha^k b}$


 * $\alpha$ is a primitive complex $n$th root of unity
 * $b$ is any complex number such that $b^n = a$.

From Euler's Identity:
 * $-1 = e^{i \pi}$

From Exponent of Product:
 * $\paren {\exp \dfrac {i \pi} n}^n = e^{i \pi}$

and so:
 * $b = \exp \dfrac {i \pi} n$

We also have by definition of the first complex $n$th root of unity, and from First Complex Root of Unity is Primitive:
 * $\alpha = \exp \dfrac {2 i \pi} n$

Hence: