 Solution

Associate to $c=(c_1,\dots ,c_n{)}^T$ the polynomial $p(t)={\sum }_{j=1}^n{c}_j{t}^{j-1}$. The $i$th component of $Ac$ is exactly $p(a_i)$. If $Ac=0$, this polynomial of degree at most $n-1$ has $n$ distinct roots. Repeated use of the factor theorem would make ${\prod }_{i=1}^n(t-a_i)$ divide any nonzero such polynomial, contradicting its degree. Hence $p$ is the zero polynomial and all $c_j=0$. The converse is immediate, proving $\ker A=\{0\}$.

A square matrix has full column rank if and only if its kernel is zero, by rank-nullity theorem. Its row rank equals its column rank. Therefore this Vandermonde matrix has row rank $n$, and its rows $v_1,\dots ,v_n$ are a basis and span ${\mathbb{R}}^n$.