Could anyone give an easy to understand proof of this fact or at least tell if the proof below is correct?
We know that every matrix is row equivalent to a reduced row echelon form. An $m\times m$ square matrix $A$ of rank $m$ is row equivalent to an identity matrix from the definition of reduced row echelon form:
1) all non-zero rows are above rows of all zeros
2) the leading coefficient is strictly to the right of the leading coefficient of the row above
3) the leading coefficient is $1$ and is the only non-zero number in its column
Plus, there's a theorem saying that the rank of any matrix equals the number of non-zero rows in its reduced row echelon form.
Now we know that $A$ has $m$ non-zero rows, then every row contains a leading coefficient $1$ that is shifted to the right with respect to the leading coefficient in the row above. The only way we can arrange $m$ 'ones' in this manner is by placing them on diagonal (we can prove it by noting the the required number of 'shifts to the right' of the leading coefficients when going one row down is $m$ - actually - is it clear enough for a proof?). And we know there is nothing except $1$ in any of those columns - we have a diagonal matrix.
Can you make it shorter without referring to some more complicated terminology?