Let A be an n x n matrix. Denote it's columns b v_{1}, v_{2}, ... v_{n}. Keep in mind these are column vectors in R^{n}.
Prove that A is invertible if and only if {v_{1}, v_{2}, ... v_{n}} spans R^{n}.
I have no idea what you know and can use. But here is a quick proof using things I feel ought to be done before proving this theorem.
$\displaystyle A$ spans $\displaystyle \mathbb{R}^n$
iff
$\displaystyle Ax=b$ is consistent for all $\displaystyle b \in \mathbb{R}^n$
iff
$\displaystyle A$ has $\displaystyle n$ pivot rows (and also $\displaystyle n$ pivot columns since $\displaystyle A$ is square)
iff
$\displaystyle A$ is row equivalent to the $\displaystyle n\times n$ identity matrix
iff
$\displaystyle A$ is invertible.
First show that if you take A times $\displaystyle \begin{bmatrix}1 \\ 0 \\ 0 \\ \cdot\cdot\cdot \\ 0\end{bmatrix}$ you get the first column of A, that if you take A times $\displaystyle \begin{bmatrix} 0 \\ 1 \\ 0 \\ \cdot\cdot\cdot \\ 0 \end{bmatrix}$ you get the second column, etc.
In other words, A maps the standard basis for R^{n} into the columns of A. Those vectors span the range of A.