Below is a simple determinant. I need to show that it is not 0, so that the corresponding matrix is invertible. $$ D = \begin{vmatrix} 0! & 1! & 2! & \ldots & x!\\ 1! & 2! & 3! & \ldots & (x+1)! \\ 2! & 3! & 4! & \ldots & (x+2)! \\ \vdots & \vdots & \vdots & \ldots & \vdots \\ y! & (y+1)! & (y+2)! & \ldots & (x+y)! \end{vmatrix} $$ Obviously, we can factor out $0!1!\ldots y!$ and get entries which are falling factorials, but I do not see how to continue.

The determinant of a similar 3X3 matrix was considered here and a stronger statement was proved on the remainder that the determinant has module 4 (after division by the obvious factors).

Below is a simple determinant. I need to show that it is not 0, so that the corresponding matrix is invertible. $$ D = \begin{vmatrix} 0! & 1! & 2! & \ldots & x!\\ 1! & 2! & 3! & \ldots & (x+1)! \\ 2! & 3! & 4! & \ldots & (x+2)! \\ \vdots & \vdots & \vdots & \ldots & \vdots \\ y! & (y+1)! & (y+2)! & \ldots & (x+y)! \end{vmatrix} $$ Remark: As pointed out in the comments, obviously we must have $y=x$ in order to have a square matrix.

Obviously, we can factor out $0!1!\ldots y!$ and get entries which are falling factorials, but I do not see how to continue.

The determinant of a similar 3X3 matrix was considered here and a stronger statement was proved on the remainder that the determinant has module 4 (after division by the obvious factors).