= Solution
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> $V$. \b[Independence plus spanning makes $S$ a <basis>.]
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