linear algebra - Determinant of $ntimes n$ matrix with parameter

Problem:

Let $\delta \in \mathbb{R}^+$ and $n\in \mathbb{N}$. The matrix $A_n = (a_{i,j}) \in \mathbb{R}^{n\times n}$ is defined as

$$a_{i,j} = \prod_{k=0}^{i-2}\left((j-1)\delta +n-k\right)$$
Prove that
$$\det A_n = \delta ^{\frac{1}{2}n(n-1)}\prod^{n-1}_{k=0}k!$$

So there is

$$A_1 = \pmatrix{ 1\\ }$$

$$A_2 = \pmatrix{ 1&1\\ 2&\delta+2\\ }$$

$$A_3 = \pmatrix{ 1&1&1\\ 3&\delta+3&2\delta+3\\ 6&(\delta+3)(\delta+2)&(2\delta+3)(2\delta+2)\\ }$$

$$\vdots$$

I eventually managed to prove it by converting the matrix to upper triangular using elementary row operations, but the proof is just too complicated, involves things like $(k−2)\delta+n−(i−(k−1))+1)$-th multiples of certain rows (for that matter it is quite long, so I am not fully including it here). So it somehow feels like not the best possible way to do this.

What are some another ways to prove this?

Answer

Finally found this in a literature, namely Calculation of some determinants using the s-shifted factorial by Jean-Marie Normand. In Lemma $1$ he proves equation $(3.5)$:

For complex numbers $z_j, s, b_i$ we have $$\tag{1}\det [(b_i+z_j)_{s;i}]_{i,j=0,\dots,n-1}=\prod_{0\leq i \leq j \leq n-1}(z_j-z_i)$$ where
$(z)_{s;i}=z(z+s)\dots(z+(n-1)s)$ is called $s$-shifted factorial and $\prod_{0\leq i\leq j\leq n-1}(z_j-z_i)$ is a Vandermonde determinant.

Furthermore Appendix B shows in equation (B$.5$) that for special case $b_i=0, z_j=b+aj$ one has

$$\det[(b_i+z_j)_{s;i}]_{i,j=0,\dots,n-1}=\det[(b+aj)_{s;i}]_{i,j=0,\dots,n-1}=a^{n(n-1)/2}\prod_{j=0}^{n-1}j!.\tag{2}$$

Before applying to our problem, we first need to re-index the range to $i,j=0,\dots,n-1$, then we have $a_{i,j}=(j\delta+n)_{1;i}$. Now choosing $b_i=0, a=\delta, b=n, s=1$ in $(2)$ gives exactly the statement we wanted to prove.