Let $\{v_1,\ldots,v_r\}\subset V$ be a set of $r$ linearly independent vectors, and let $\{w_1,\ldots,w_n\}$ be a basis for $V$. Then there exist $\alpha_{ij}$ with $i\in\{1,\ldots,r\}$ and $j\in\{1,\ldots,n\}$ such that $v_i=\sum_{j=1}^n \alpha_{ij}w_j$. Consider the system of equations: $$\sum_{i=1}^r \alpha_{ij} x_i=0, \ j\in\{1,\ldots,n\}$$
Let $(x_1,\ldots,x_n)$ be a solution. Then $$\sum_{i=1}^r x_i v_i=\sum_{i=1}^r\sum_{j=1}^n\alpha_{ij}x_i w_j=\sum_{j=1}^n\left(\sum_{i=1}^r\alpha_{ij}x_i\right)w_j=\sum_{j=1}^n 0\cdot w_j=0$$
$\{v_1,\ldots,v_r\}$ is linearly independent, so it must be $(x_1,\ldots,x_n)=0$. Hence, the previous system of equations either has unique solution or hasn't solution at all. Anyways, this means that it must have at least as many equations as variables, i.e. $r\leq n$.
Then, any linearly independent set must have, at most, $n$ vectors. Let now $\{v_1,\ldots,v_n\}$ be a linearly independent set and let's prove it must span $V$. Assume it doesn't, and take $v\in V\setminus\operatorname{Span}(v_1,\ldots,v_n)$. Then $\{v,v_1,\ldots,v_n\}$ is a linearly independent set with $n+1$ elements; a contradiction. Hence $\{v_1,\ldots,v_n\}$ spans $V$ so it's a basis.
