Cyclotomic integers: Why do we have $x^n+y^n=(x+y)(x+\zeta y)\dots (x+\zeta ^{n-1}y)$? Why do we have the factorization $$x^n+y^n=(x+y)(x+\zeta y)\dots (x+\zeta ^{n-1}y)$$ for $\zeta$ a primitive $n^{\text{th}}$ root of unity where $n$ is an odd prime?

If $y = 0$, the factorisation reads $x^n = \underbrace{x\,.x \dots x}_{n\ \text{times}}$.
Consider $y\neq 0$. If $\zeta \in \mathbb{C}$ is a primitive $n^{\text{th}}$ root of unity then $\zeta^n = 1$ and $\zeta^k \neq 1$ for $k = 1, \dots, n-1$. In particular, $\zeta, \zeta^2, \dots, \zeta^{n-1}, \zeta^n=1$ are all distinct. Now consider $x^n + y^n$ as a polynomial in $x$ over the complex numbers (treat $y$ as a fixed constant). Then by the Fundamental Theorem of Algebra, $x^n + y^n$ has precisely $n$ zeroes. For each $k = 1, \dots, n$, $x = -\zeta^ky$ is a zero as 
\begin{align*}
x^n + y^n &= (-\zeta^ky)^n + y^n\\ 
&= (-1)^n(\zeta^k)^ny^n + y^n\\ 
&= -(\zeta^n)^ky^n + y^n\qquad \text{as $n$ is odd, $(-1)^n = -1$}\\ 
&= -y^n + y^n\\ 
&= 0.
\end{align*} 
As $y \neq 0$, $x = -\zeta^ky$ for $k=1, \dots, n$ are all distinct, so they are all of the zeroes of $x^n + y^n$. By the factor theorem, each zero corresponds to a linear factor, so we have 
\begin{align*}
x^n + y^n &= a(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y)(x+\zeta^ny)\\
&= a(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y)(x+y)\\
&= a(x+y)(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y)
\end{align*}
for some $a \in \mathbb{C}$. Comparing the coefficients of $x^n$, we deduce that $a = 1$ and arrive at the desired factorisation
$$x^n+y^n = (x+y)(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y).$$
