It was proved on page 26 of this note the following result:
Let $\xi$ be an algebraic number that is not a root of unity, then there exists an $n_0\geq 0$ with the property that $$\beta=\sum_{j=-n^2}^{-n}c_j\xi^j=\sum_{j=n}^{N}c_j\xi^j, \forall j,~c_j\in\mathbb{Z}$$ for some $N\geq n\geq n_0$ and $|c_j|\leq |j|^{20}$ implies that $\beta=0$.
Surprising, the proof given in this note uses the valuation on the field $\mathbb{Q}(\xi)$. So, I am wondering could anyone gives another proof of this result?
Thanks in advance!
Remarks:
1, Intuitively, by considering the minimal polynomial of $\xi$, say $\xi^k=a_0+a_1\xi+\cdots+a_{k-1}\xi^{k-1}$, we could simplify $\beta$ as a linear combination of terms $1,\xi,\cdots,\xi^{k-1}$, then two expressions would give us some equality of the corresponding coefficients, which might be quite large in absolute value, then we can expect to use the restriction on $c_j$ to get $\beta=0$. But it is not easy for me to figure out the details, and I do not think this approach really works..
2, I do not know which tags or even the title best describe this question, so feel free to modify them.