Does a set of $n$ independent vectors in $R^{n}$ always span $R^{n}$? Does a set of $n$ independent vectors in $R^{n}$ always span  $R^{n}$?
If so, what's the reasoning behind this?
 A: Note that any set of independent vectors can always be extended to be a basis (i.e. independent spanning set) of the vector space.
Now, $\mathbb{R^n}$ has dimension $n$- which means that any basis can have no more than $n$ elements (in fact a basis has exactly $n$ elements). Now if your linearly ind. set of $n$ vectors didn't span $\mathbb{R^n}$ then we would have a basis consisting of ore than $n$ elements by extending the set of l.i. vectors.
So, any set of $n$ linearly ind. vectors in $\mathbb{R^n}$ spans $\mathbb{R^n}$.
A: In Linear Algebra, I learned a theorem which our professors called "Third One Free" (free translation). If I remember it correctly (and I'm not sure I do), it states that given a set of vectors $B$ in a vector space $V$ - we say that if two of these three are correct, then so is the third:


*

*$span(B)=V$ (i.e. $B$ spans $V$).

*$B$ is linearly independent.

*$|B|=dim(V)$


So, in that case, we have the 2nd and the 3rd statements are correct (size and linear independence), hence, according to the "Third One Free" theorem, the 1st statement (the vector set spans the space) is correct.
Hope I helped, and that my English wasn't too horrible :) 
A: This is indeed very easy if you already know what dimension is and, more specifically, why the definition of dimension is correct. The other answers already show how to use dimension to solve your problem: basically, the $n$ independent vectors span something $n$-dimensional, and if they didn't span all of $\mathbb{R}^n$, then $\dim \mathbb{R}^n$ would be larger than $n$, a contradiction, and we are done.
However, it is a good exercise to prove your statement without using the notion of "dimension" at all. Here is a plan:


*

*Your statement is equivalent to the following: there cannot exist $n+1$ linearly independent vectors in $\mathbb{R}^n$. (You should make sure that you understand why this is indeed equilavent to your question).

*Assume there exist $n+1$ linearly independent vectors $v_1, \ldots, v_{n+1} \in \mathbb{R}^n$. Each vector is a row with $n$ real entries. Let's write down all the $n+1$ vectors under each other. We get an $(n+1) \times n$ matrix whose rows are linearly independent.

*We can do some row transformations of our matrix. You should check that row transformations do not affect linear dependence, i.e. any row transformations will leave all the $n+1$ rows of the matrix linearly independent.

*Using a Gauss-like algorithm it is possible to reach a state when one of the rows in the matrix is zero (because the number of rows is strictly larger than the number of columns).

*If one of the rows is zero, then the rows are linearly dependent. But we know that row operations leave the rows linearly indepednent, so this is a contradiction. Done.


PS: maybe your source (book/class) teaches to think of elements of $\mathbb{R}^n$ as columns rather than rows. This is not a problem, just "transpose" the whole argument.
PPS: of course, I assumed that by $R$ you mean $\mathbb{R}$ (the field of real numbers).
PPPS: you can also do this without matrices (in case you don't like them). I'm too lazy to write down that argument too, maybe someone will do it in another answer.
