Set of $n$ vectors spanning an $n$ - dim vector space implies set is a basis 
Theorem:
If a set of n vectors spans an n-dimensional vector space, then the set is a basis for that vector space.

Attempt:
Let S be a set of n vectors spanning an n-dimensional vector space.
This implies that any vector in the vector space $\left(V, R^{n}\right)$ is a linear combination of vectors in the set S.
It suffice to show that S is linearly independent.
Suppose that S is linearly dependent.
Then for some vectors $\vec{v}_{i}, \exists i \in \mathbb{Z}_{1}^{n}$ in S that may be expressed as a linear combination of some vectors in S, the removal of $\vec{v}_{i}$ does not affect the span of set S.
Any hints to bring me forward is highly appreciated.
Thanks in advance.
 A: Let $S=\{x_1,x_2,\ldots,x_n\}$ be a set spanning the vector space $V$ of dimension $n$. Suppose $S$ is not a basis of $V$. Then $S$ is linearly dependent. Thus there exists $x_i\in S$ such that $x_i$ is a linear combination of remaining $n-1$ vectors as $S\setminus \{x_i\}$ also spans $V$. Thus $n-1$ vectors can span $V$, which is a contradiction. Hence $S$ is  a basis. 
A: It's impossible to say what's a correct solution without knowing what previously proved results you're allowed to use (for example, the other answer is fine if you're given that $n-1$ vectors cannot span $V$.)
Since the word "dimension" appears it seems reasonable to conjecture that you've already shown that any basis for $V$ must have exactly $n$ elements; this is required before the typical definition of "dimension" makes sense. Assuming that it's an easy exercise:
Show that $S$ contains a minimal spanning set: There exists $S_1\subset S$ such that $S_1$ spans $V$ and if $S_2\subset S_1$ and $S_2$ spans $V$ then $S_2=S_1$. Show that it follows that $S_1$ is independent.
So $S_1$ is a basis. Hence that previous result show that $S_1$ contains exactly $n$ elements; since $S_1\subset S$ this shows that $S_1=S$.
