TITLE: Prove that a subspace of dimension $n$ of a vector space of dimension $n$ is the whole space.
QUESTION [7 upvotes]: Maybe this is a stupid question.
I was brought to this from the observation that an infinite dimensional vector space can have proper subspace that have the same dimension of the whole space.
But, for a finite dimensional vector space, as $\mathbb{R}^n$ it seems obvious that, if $S$ is a subspace of dimension $n$,  than  $S=\mathbb{R}^n$. 
I've tempted to give a proof this statement, thinking that was a simple exercise, but I am no able to find  a good way.
I suppose that I'm lost in a teacup so I need an help.

Added after the first answers.
The answer of @Mark Joshi was my first idea, so I llustrate better where is my trouble.
Suppose $S \in \mathbb{R}^n$ be a subspace of dimension $n$. This means, by definition, that:
1) there exists a set of linearly independent vectors $\{v_1, v_2, \cdots, v_n\}$ with $v_i \in S$
and
2) $\forall v \in S$ we have $v \ne v_i \; \forall i \Rightarrow \{v, v_1,\cdots,v_n\} $ are linearly dependent.
Note that 2) is valid for $v\in S$ and all $v_i \in S$.
Since $\mathbb{R}^n$ is a vector space, we know that there exists a set $\{u_1,u_2,\cdots,u_n\}$ of vectors $u_i \in \mathbb{R}^n $ such that:
3) $\{u_1,u_2,\cdots,u_n\}$ are linearly independent
and
4) $\forall u \in \mathbb{R}^n$ we have $u \ne u_i \; \forall i \Rightarrow \{u, u_1,\cdots,u_n\} $ are linearly dependent.
Now, if $u, u_i \in \mathbb{R}^n/S$ I can not conclude from 1) and 2) that $u$ is a linear combination of $\{v_i\}$ and this means that I cannot decide if it is in the span of $ \{v, v_1,\cdots,v_n\} $. Surely it is a linear combination of $\{u_1,u_2,\cdots,u_n\}$ , but some (or all) of these $u_i$ can well not be elements of $S$.
So it seams that we can have a vector $u$ that is a linear combination of $n$ linearly independent $\{u_i\}$ but we connot proof that it is also a linear combination of $\{v_i\}$. And this is my trouble.
It seems to me that the way suggested by @Awllower has the some problem when we search to show that $f :S \rightarrow \mathbb{R}^n$ defined by the correspondence $v_i \rightarrow u_i$ is bijective, because we have to proof that $\forall u \in \mathbb{R}^n$ there exists $v \in S$ such that $v=\sum_i a_i v_i \Rightarrow u=\sum_i a_i u_i$ and this come to the same trouble.
At last, I am sure that there is something of logically wrong in my reasoning, but it is so subtle or so stupid that i cannot see it.

REPLY [10 votes]: suppose we have a subspace of dimension then it has a basis of $n$ linearly independent vectors $v_1, v_2, \dots, v_n.$ 
If we have a vector $v \in R^n$ either it is in their span or it is not. if it is not then we have $n+1$ linearly independent vectors which is impossible since the dimension of $R^n$ is $n.$