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.?

Let $n$ be odd. The $n$ distinct roots of the polynomial $X^n-1$ are $1,\zeta,\dots,\zeta^{n-1}$
$$
X^n-1=(X-1)(X-\zeta)\cdots(X-\zeta^{n-1})
$$
Apply the transformation $X\mapsto -\frac{x}{y}$
$$
\left(-\frac{x}{y}\right)^n-1=\left(-\frac{x}{y}-1\right)\left(-\frac{x}{y}-\zeta\right)\cdots\left(-\frac{x}{y}-\zeta^{n-1}\right)
$$
Finally, multiply through by $-y^n$
$$
x^n+y^n=(x+y)\cdots(x+\zeta^{n-1}y)
$$
