Way to show $x^n + y^n = z^n$ factorises as $(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n$ For odd $n$ the Fermat equation $x^n + y^n = z^n$ factorises as $$(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n,$$
where $\zeta = e^{2 \pi i/n}$. I tried seeing this was true by multiplying the factors out by hand but got into a mess. Is there a neat way of showing this (e.g. is there a useful identity that can be used involving products of the $\zeta^i$ etc.?

We will make use of the following theorem

Theorem Let $P(x)$ be a polynomial and $P(z)=0$, then $x-z$ divides $P(x)$.

Let $n$ be odd and write $$P(x) = x^n + y^n,$$ then $P(-\zeta^r y)=0$ for all $n$ powers of $r$.
Therefore $(x+y)(x+\zeta y)(x+\zeta^2 y)\cdots$ divides $P(x)$ but it has the same degree so its equal (up to a constant factor which is easily seen to be 1).
