Can someone explain this part of this property of invertible matrices proof? The property states, "A square matrix A is invertible iff it can be written as the product of elementary matrices"
I'm confused on the part of the theorem where they're trying to show that if A is invertible, then it can be written as the product of elementary matrices.
This is that section of the proof:
"Assume A is invertible. You know the system of linear equations represented by Ax=0 has only the trivial solution. But this implies that the augmented matrix [A 0] can be rewritten in the form [I 0] (using elementary row operations corresponding to E1,E2,...,Ek). So, Ek,...,E2,E1A I and it follows that A = E1-1E2-1...Ek-1 . A can be written as the product of elementary matrices."
I just don't get how knowing that Ax=0 has only the trivial solution implies that [A 0] can be written in the form [I 0]. Wasn't it already obvious that A can be rewritten as I since it's invertible? And obviously if there's a 0 matrix adjoined A to it it's going to stay the zero matrix no matter what row operations are done on it? What's the point of doing that?
I'm just generally confused on this proof

If the system $Ax=0$ hasn't a unique solution, then it has infinitely many (it cannot have no solution, because the zero vector is always a solution). In this case, row operations end by giving a matrix having its last line(s) entirely filled of $0$ and you can't obtain $[I\; 0]$. 
Now, if $Ax=0$ has a unique solution, then this solution is necessarily the zero vector. Hence, the system is equivalent to $x=0$ (or $Ix=0$). This means exactly that $[A\; 0]$ can be transformed (via elementary row operations) into $[I\; 0]$. 

Wasn't it already obvious that A can be rewritten as I since it's invertible? 

It was obvious that there is a matrix $P$ s.t. $PA=I$ but it wasn't obvious at all that there is some $P$ being a product of elementary row matrices s.t. $PA=I$ (this could have existed for only certain particular kind of invertible matrices). 
