Matrix/Staircase form by elementary manipulations/Fact

Let ${}K$ be a field, and let ${}M$ denote an ${}m\times n$-matrix over ${}K$.

Then there exist

elementary row operations, and a (new) numbering of the columns

$j_{1},\,j_{2},\ldots ,j_{n},$

and an ${}r\leq n$ such that, in the new matrix, the columns have the form

$s_{j_{k}}={\begin{pmatrix}b_{1,j_{k}}\\\vdots \\b_{k,j_{k}}\\0\\\vdots \\0\end{pmatrix}}{\text{ with }}b_{k,j_{k}}\neq 0{\text{ for }}k\leq r$

and

$s_{j_{k}}={\begin{pmatrix}b_{1,j_{k}}\\\vdots \\b_{r,j_{k}}\\0\\\vdots \\0\end{pmatrix}}{\text{ for }}k>r.$

By further elementary row operations, and by swapping of columns, the matrix can be brought to the form

${\begin{pmatrix}d_{11}&*&\cdots &*&*&\cdots &*\\0&d_{22}&\cdots &*&*&\cdots &*\\\vdots &\vdots &\ddots &\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &d_{rr}&*&\cdots &*\\0&0&\cdots &0&0&\cdots &0\\\vdots &\vdots &\ddots &\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &0&0&\cdots &0\end{pmatrix}}$

with

${}d_{ii}\neq 0$.

Write a proof