= Solution
Associate to $c=(c_1,\ldots,c_n)^T$ the <polynomial> $p(t)=\sum_{j=1}^nc_jt^{j-1}$. 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 \b[$\ker A=\{0\}$].
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 $v_1,\ldots,v_n$ are a <basis> and \b[span $\mathbb R^n$].
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