Let $A$ be an $m\times n$ matrix. Show that $\mathrm{rank}(A)=n$ iff $\mathrm{rref}A$ has no zero columns, using only the definition of rank and reduced row echelon form ($\mathrm{rref}$).
My attempt: If $\mathrm{rref}A$ has no zero columns, suppose $\mathrm{rank}(A)< n$. For every row $i=1,\ldots,k$, let $j_i$ be the column in which he first non-zero entry of row $i$ appears. Note that $$\{j_1,...j_k\}\subset\{1,...,n\}.$$ This implies that there is at least one column $j_i$ in which there isn't a first non-zero entry of any row $i$. The problem is that I haven't proved that the previously mentioned column is a zero column.
It is much easier to think of this in terms of rows.
By definition, the rank is the number of pivots in the rref of $A$ (your definition may differ). However, every row can have at most one pivot (by the definition of rref). If a row does not contain a pivot, then (again, by the definition of rref) it is a pivot row.
So, suppose that the rank of $A$ is $n$. There are $n$ rows, so every row must have a pivot. So, there are no zero rows in the rref.
Conversely suppose that $A$ has a zero row. Then there is a row that has no pivot. So, there must be less than $n$ pivots overall, which means that the rank is less than $n$.