If A is an invertible nxn matrix, $A\bar x = \bar 0$ has only the trivial solution of $\bar 0$ because $$A^{-1}A\bar x = \bar x = A^{-1}\bar0 =\bar0$$ This implies A is a product of elementary matrices because it implies A is row equivalent to I

Let's assume the A is in REF, if one of the diagonal elements were $0$ we would have a row filled with $0$'s and would have more unknowns than equations (proof n>m implies nontrivial solution omitted) and as a consequence has a nontrivial solution. Therefore, I is the RREF of A and A is composed of elementary row operations of I

If A is an invertible nxn matrix, $A\bar x = \bar 0$ has only the trivial solution of $\bar 0$ because $$A^{-1}A\bar x = \bar x = A^{-1}\bar0 =\bar0 \tag{1}\label{1}$$ This implies A is a product of elementary matrices because it implies A is row equivalent to I

Let's assume the A is in REF, if one of the diagonal elements were $0$ we would have a row filled with $0$'s and would have more unknowns than equations (proof n>m implies nontrivial solution omitted) and as a consequence has a nontrivial solution. Therefore, since we have no nontrivial solution $\ref{1}$, I is the RREF of A and A is composed of elementary row operations of I