Basis of $\mathbb R$ seen as a $\mathbb Q$-vector space

Question

$\mathbb R$ is clearly a $\mathbb Q$-vector space since all axioms needed to be a vector space are verified. Its basis is infinite because it would need to have all roots, $\pi$, $e$...

1. Is there any quick and "relatively easy" but solid proof that this statement is true ? I don't know how to formally write "the basis needs to contain all (actually not all but still a lot) irrationals and |$\mathbb R$ \ $\mathbb Q$| would be bigger than some uncountable set.
2. Can we write the basis in a simple form ? What I mean is that, the basis of polynomials $\mathbb R$[t] is infinite but can be written as {$t^k$ , k $\in$ $\mathbb N$}

Answer 1

1. By definition, the powers of $\pi$ are linearly independent over $\mathbb Q$ since $\pi$ is transcendental. Therefore, $\mathbb Q[\pi]$ is a infinite-dimensional subspace of $\mathbb R$.

2. No description of a basis for $\mathbb R$ over $\mathbb Q$ is known. In most if not all definitions, a description is a string in a language and so there are only countably many descriptions, not enough to describe an uncountable set.