Total nonnegativity of the flipped excedance matrix

Let $M(m,n)$ be the excedance matrix, and let $N(m,n)$ be obtained by reversing the order of its rows: its $i$th row is the $(m-i+1)$st row of $M(m,n)$. A matrix is totally nonnegative if every square submatrix has nonnegative determinant. Total-nonnegativity conjecture. The matrix $N(m,n)$ is totally nonnegative. This conjecture concerns the sign and minor structure of the excedance matrix and is presented without a resolution in the supplied text.