I am looking for a proof of the claim that given a non-singular matrix $A$ that there exists a row permutation matrix $P$ such that $PA$ has a fully populated diagonal.
(Or in the case of a potentially singular $A$ that if after an optimal permutation we end up with $n$ non-zeros along the diagonal that $\textrm{rank}\, A \le n$.
The determinant can give a proof.
A matrix $A=(a_{ij})$ is non-singular iff its determinant is nonzero.
The determinant of $A$ is the signed sum of products $\prod_i a_{i,\sigma(i)}$ for all possible permutations $\sigma$.
Having no row permutation which produces only nonzero diagonal elements means exactly that all these products are zero, but then $\det A=0$ by the above.
The same can be applied for the case $\mathrm{rank\,}A\ge n$, once we know that this means that $A$ has an $n\times n$ submatrix with nonzero determinant.