Linearly independent numerous problem [duplicate]
$x^n-2$ is an Eisenstein polynomial so is irreducible. Thus $\mathbb Q(\sqrt[n]2)=\mathbb Q[x]/(x^n-2)$ is a degree $n$ extension of $\mathbb Q$. In particular, $1,\sqrt[n]2,\sqrt[n]2^2,\dots,\sqrt[n]2^{n-1}$ are $\mathbb Q$-linearly independent.