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Theorem on linear independence

How would one go about proving the second part of the if and only if theorem in the above link? I have proved the first half that is if S is linearly independent it cannot have any vector be spanned by the vectors listed before it,but I can't find a logical reason to prove the converse of this(ie. if for all vectors v in S , if v is not spanned by the vectors listed before it , it is linearly independent).

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    This isn't a theorem, it's a definition. You don't prove it, you accept it as the meaning of the words "linearly independent". – 2017-02-14
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    But I am unable to relate this statement or definition with other definitions., which are more general than this – 2017-02-14
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    What other definitions would you like to relate it to? – 2017-02-14
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    @ziggurism It can be used as definition, but I think that here it is a theorem after all. It might be that it must be proved on base of a definition like: "$v_1,\dots,v_n$ are linearly independent if the equality $\lambda_1v_1+\cdots+\lambda_nv_n=0$ implies that $\lambda_i=0$ for every $i$" – 2017-02-14
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    @drhab , Right! [link] (https://i.stack.imgur.com/4cfmM.png) , I am asked to prove this as well. – 2017-02-14

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Hint: use induction on $k$.

Induction step:

Suppose that $v_1,\dots,v_{k-1}$ are linearly independent and that $v_k\notin\text{span}(\{v_1,\dots,v_{k-1}\})$. Now try to prove that under these conditions also $v_1,\dots,v_{k-1},v_k$ are linearly independent.