Let $p$ be a prime number $\ge2$ and $u = \cos\left(\frac 2p\pi\right)+i\sin\left(\frac2p\pi\right) \in \mathbb C$. Prove that $u$ is root of $f(x)=x^{p-1}+x^{p}+...+x+1$.
I know that $f(x)$ is irreducible over $\mathbb Q$ by the criterion of Eisenstein and $u^p=1$.
The indicated number is a complex $p$th root of $1$. The polynomial $x^p-1$ factors as $$x^p-1=(x-1)(x^{p-1}+x^{p-2}+\cdots+1)$$ Since the number is a root of this polynomial but is not equal to $1$, it follows that it is a root of the latter polynomial in the product, which is what we wanted to show.