Recall that a vector $w$ is a linear combination of $w_1,\ldots,w_k$ if there exist scalars $\lambda_1,\ldots,\lambda_k$ such that $w = \lambda_1w_1 + \cdots + \lambda_kw_k.$ A set $\{w_1,\ldots,w_k\}$ is linearly dependent if there exists a non-trivial solution to $\lambda_1w_1+\cdots+\lambda_kw_k = 0$, i.e. not all $\lambda_i = 0.$
Assume that $v_1 \neq 0$ and that $\{v_1,\ldots,v_n\}$ is linearly dependent.
Since $v_1 \neq 0$ it follows that $\{v_1\}$ is linearly independent.
Next, consider $\{v_1,v_2\}$. Either $\{v_1,v_2\}$ is linearly independent (LI) or linearly dependent (LD). If $\{v_1,v_2\}$ is LD then you have your $v_j$, namely $j=2$, because $\{v_1\}$ was LI while $\{v_1,v_2\}$ is LD meaning that $v_2$ is a linear combination of $v_1$.
If $\{v_1,v_2\}$ is LI then consider $\{v_1,v_2,v_3\}$. Either $\{v_1,v_2,v_3\}$ is LI or LD. If $\{v_1,v_2,v_3\}$ is LD then you have your $v_j$, namely $j = 3$, because $\{v_1,v_2\}$ was LI while $\{v_1,v_2,v_3\}$ is LD meaning that $v_3$ is a linear combination of $v_1$ and $v_2$.
If $\{v_1,v_2,v_3\}$ is LI then consider $\{v_1,v_2,v_3,v_4\}$, etc.
Continue this process. You will eventually find a $v_j$ for which $\{v_1,\ldots,v_{j-1}\}$ is LI and $\{v_1,\ldots,v_j\}$ is LD meaning that $v_j$ is a linear combination of $v_1,\ldots,v_{j-1}.$ If you don't find such a $v_j$ then you have a contradiction: you would have that $\{v_1,\ldots,v_n\}$ was LI, but you assumed it was LD!