Linearly independent set in $\mathbb R$ over $\mathbb Q$ The set $$\left\{\sqrt[2^n]{2},\quad n\in\mathbb N\right\}\subset\mathbb R$$ is linearly independent over $\mathbb Q$.
Is there an easy or elementary way to see this? I.e., why does the equation $\alpha_1 2^{2^{-i_1}}+\cdots +\alpha_N 2^{2^{-i_N}}=0$ not have a non-trivial solution in $\alpha_i\in\mathbb Q$ for any choice of $N\in\mathbb N$ and $i_1,\dots,i_N\in\mathbb N$?

More generally, for each $N \in \mathbb N$, the set $\left\{\sqrt[2^n]{2} : n=0,\dots,N\right\}$ is linearly independent because $X^{2^N}-2$ is irreducible over $\mathbb Q$ due to Eisenstein's criterion.
