Given finite field $GF(2^b)$ with elements $v_1,\ldots,v_{2^b}$, and integer $a$ such that $2^{b} \geq 2(a+1)!$, I am trying to locate a proof that the set of column vectors $e_i = (1,v_i,v_i^3,\ldots,v_i^{2a+1})^{\operatorname{T}}$, $1 \leq i \leq 2^b$, form a spanning set for additive group $\mathbb{Z}_2^{1+b(a+1)}$ (where each $v_i$ is itself a column vector, so $e_i$ is a vector of length $1+b(a+1)$).
Firstly, is this statement actually true? I'm aware a series of vectors don't really 'span' a group in the sense of a basis, but I mean that each element of $\mathbb{Z}_2^{1+b(a+1)}$ is a linear combination of the $e_i$.
Secondly, if indeed it is true (I believe/hope it is!), is there a particularly nice and elementary or concise way to prove it? Thanks for your help!