In Lay's Linear Algebra and Its applications, he gives a proof the uniqueness of the reduced echelon form.
Consider two row equivalent matrices U and V in reduced echelon form. He says that the pivot columns of U and V are precisely the nonzero columns that are not linearly dependent to the columns to their left. Since U and V are row equivalent, their columns have the same linear dependence relations. Hence, the pivot columns of U and V appear in the same locations.
I'm having a hard time seeing how the bolded statement follows from the previous two statements.
Suppose that the $k$-th column of $U$ is a pivot column; then it is linearly independent of the columns to its left. If the $k$-th column of $V$ were not linearly independent of the columns to its left, it would be linear combination of them, and the first $k$ columns of $V$ would therefore satisfy a linear dependence relation that the first $k$ columns of $U$ do not satisfy. But that is impossible, because $U$ and $V$ are row-equivalent, and their columns therefore satisfy the same linear independence relations: if the first $k$ columns of $V$ are linearly dependent, then so are the first $k$ columns of $U$, and vice versa.
Therefore the $k$-th column of $V$ must be linearly independent of the columns to its left and hence be a pivot column. This shows that every pivot column of $U$ must be a pivot column of $V$, and the same argument with the rôles of $U$ and $V$ reversed shows that every pivot column of $V$ must be a pivot column of $U$. It follows that $U$ and $V$ must have exactly the same pivot columns.