Suppose $w$ is $p$-th root of unity. Why when from $$a_o + a_1 \omega + a_2 \omega^2 ... + a_{p-1} \omega^{p-1} = 0$$ we can conclude that $$a_0 = a_1 = a_2 = ... = a_{p-1}.$$ And later use it to find some nonzero $a_i.$
I have found this piece in this answer.
I thought to multiply by $w$ and so change the roles of $a_0$ and $a_{p-1},$ and so on. But i also think that all the coefficients must be zeroes.
EDIT: $p$ is prime.
Consider the map $\mathbb Q$-linear map $\mathbb Q^p \to \mathbb Q[\omega], (a_0, \dotsc, a_{p-1}) \mapsto a_0 + a_1 \omega + \dotsb a_{p-1}\omega^{p-1}$.
Since $1, \omega, \dotsc, \omega^{p-2}$ are linear independent (the minimal polynomial has degree $p-1$), we get that the image has dimension $p-1$, hence the kernel is one-dimensional.
Clearly $(1,1, \dotsc, 1)$ is an element of the kernel, thus any element of the kernel has the form $a \cdot (1,1, \dotsc, 1)$.
Here are some more details:
We have $1+ \omega + \omega^2 + \dotsb + \omega^{p-1} = \frac{\omega^p-1}{\omega - 1}=0$, hence $(1,1, \dotsc, 1)$ is an element of the kernel. In other words, $\omega$ is a root of the polynomial $f=x^{p-1}+x^{p-2} + \dotsb + x +1$. This polynomial is irreducible, hence it is the polynomial of smallest degree, that has $\omega$ as a root. In particular $1, \omega, \dotsc, \omega^{p-2}$ are linear independent, since any linear combination would give rise to a polynomial of degree $\leq p-2$.
