I am reading the following proof about the field extension $\mathbb{C}/\mathbb{Q}$, that this extension is infinite. 

It is sufficient to show that, for no matter how big natural number $n$, there are $n$ complex numbers, that are $\mathbb{Q}$-linearly independent. 

For that, we consider the polynomial $f(x)=x^n-2$ and let $z\in \mathbb{C}$ one of its roots, e.g., $z=\sqrt[n]{2}\left (\cos \frac{2\pi }{n}+i\sin \frac{2\pi }{n}\right )$. 

We claim that the complex numbers $1, z, z^2, \dots , z^{n-1}$ are $\mathbb{Q}$-linearly independent. Indeed, in other case, there would be rationals $c\_0, c\_1, \dots , c\_{n-1}$, not all zero, such that $c\_0+c\_1z+\dots +c\_{n-1}z^{n-1}=0$, i.e., $g(z)=0$, where $g(x)=c\_0+c\_1x+\dots +c\_{n-1}x^{n-1}$. Therefore, the non-zero polynomials $f(x)$ and $g(x)$ would have $z$ as a common root, so we would conclude that $f(x)$ and $g(x)$ are not coprime. But $f(x)$ is irreducible, so we would have to conclude that $f(x)\mid g(x)$. That is not possible, because the non-zero polynomial $g(x)$ has degree smaller than the degree of $g(x)$. 

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We have that $f(x)$ and $g(x)$ have a common root, $z$. Why do we conclude that $f(x)\mid g(x)$ ? 

How have we shown in that way that the field extension is infinite? Is it because we take $n\in \mathbb{N}$ ?