Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Digital systems, linear algebra, introduction to formal proof / 4 / a / ii

Suppose a vector $v$ lies outside the span of $S$. Then $S\cup \{v\}$ is linearly independent: in an equation $\alpha v+{\sum }_i{\beta }_i{s}_i=0$, a nonzero $\alpha$ would express $v$ in that span, and with $\alpha =0$ the independence of $S$ forces every ${\beta }_i=0$.

This would give $n+1$ linearly independent vectors in a space with an $n$-vector basis, contrary to the supplied claim. Therefore $S$ spans $V$. Independence plus spanning makes $S$ a basis.