Determinant with factorials is not 0?

Question

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).

Answer 1

This is the Hankel determinant associated to the sequence $m_n = \mathbb{E}(X^n) = n!$ of moments of an exponential distribution with mean $1$. Some general results can be used to show that the sequence of Hankel determinants associated to the moments of a random variable are always positive iff the induced measure on $\mathbb{R}$ has infinite support, and some more general results can be used to exactly calculate the Hankel determinants as in the comments by calculating an appropriate sequence of orthogonal polynomials. The relevant orthogonal polynomials for the exponential distribution are the Laguerre polynomials.

I gather this is very classical material but I don't know a reference; you can see a writeup in slightly unusual language here.

Edit: Ah, here's a reference: Section 2.7 of Krattenthaler's Advanced Determinant Calculus.

Answer 2

Let's write a more general matrix $M_n(t)$ of size $n\times n$ having entries $(i+j-2+t)!$ for $1\leq i,j\leq n$. Then, we claim that $$\det(M_n(t))=\prod_{k=1}^n(k-1)!\,(k-1+t)! \tag1$$ which gives your desired determinant by setting $t=0$ and $n=x+1=y+1$.

The proof for (1) can be given by the so-called Dodgson's Condensation with runs recursively. I can supply this if needed.

Answer 3

A direct proof of T. Amdeberhan's identity (1) is as follows: we have $(i+j+t)!=(i+t)! f_j(i)$, where $f_j(x)=(x+1)(x+2)\ldots (x+j)$. Thus $$ \det ((i+j+t)!)_{0\leqslant i,j\leqslant n-1}=\prod_{i=0}^{n-1} (i+t)!\times \det (f_j(i))_{0\leqslant i,j\leqslant n-1}, $$ and since $f_j$ is a monic polynomial of degree $j$ we get for arbitrary numbers $x_0,\ldots,x_{n-1}$ $$ \det (f_j(x_i))_{0\leqslant i,j\leqslant n-1}= \det (x_i^{j-1})_{0\leqslant i,j\leqslant n-1}=\prod_{i<j} (x_j-x_i), $$ where the first equality follows from consecutive subtraction from the $j$-th column ($j=0,1,\ldots,n-1$) an appropriate linear combination of the previous columns; and the second equality is Vandermonde determinant.