Why is a matrix invertible if it can be written as the product of elementary matrices? I'm just starting to learn linear algebra and I understand Gaussion elimination but I don't understand this proof:

Is the heart of this the fact that all the while when I've been row-reducing matrices to reduced row-echelon form... I've actually been multiplying a matrix by a series of a elementary matrices?
What does the notation $[A\ \ 0]$ mean?
How is the equation $A = E_1^{-1}...E_k^{-1}$ arrived at?
 A: $[A\,0]$ is so-called block matrix notation, where a large matrix is written by putting smaller matrices ("blocks") next to one another (or above one another). In this case it means the matrix you get if you put $A$ right next to a (suitably tall) $0$ column.
Yes, every operation you've done when row-reducing a matrix may be realized by multiplying by a suitable elementary matrix.
For the final question, we start with
$$
E_k\cdots E_1A=I
$$
Then we multiply both sides from the left by $E_k^{-1}$ to get
$$
E_{k-1}\cdots E_1A=E_k^{-1}
$$
Then we multiply both sides from the left by $E_{k-1}^{-1}$ to get ... (and so on, until finally) Then we multiply both sides from the left by $E_1^{-1}$ to get
$$
A=E_1^{-1}\cdots E_k^{-1}
$$
A: When we apply an elementary row operation $E$ to a matrix $A$, what we actually have is $EA$ algebraically. So when we apply a sequence of elementary row operations to $A$, it means multiplying them by $A$ as you said.
The notation $[A\ |\ 0]$ or $[A\ \ 0]$ is equivalent to $$ \left[
\begin{array}{cc|c}
  1&2&0\\
  3&4&0
\end{array}
\right] $$ 
where $A = \begin{bmatrix}1&2\\3&4\\ \end{bmatrix}$; and the result is called "augmented matrix" (This is an example of course).
The equation $A=E_1^{-1}E_2^{-1}\cdots E_k^{-1}$ can be obtained by multiplying both sides of the equation $E_k\cdots E_2E_1A=I$ by $E_k^{-1}$, ..., $E_1^{-1}$ from the left in order.
A: "I've actually been multiplying a matrix by a series of a elementary matrices?" Yes.
"What does the notation [A0] mean?" It means augmenting a matrix A by the matrix O. If you have a matrix $((1,1),(1,1))$ and you augment it by the matrix $((2,2))$, then you get $((1,1),(1,1),(2,2))$. It is the core of Gaussian elimination (written in a more elegant way).
"How is the equation A=E(1−1)...E(k−1) arrived at?" Multiply by the inverse of the k elementary matrices (they exist and you can prove it) the previous equation. 
