"Almost Hankelized" numerical Vandermonde

Question

One of the more utilized determinant is that of Vandermonde's

$$\begin{vmatrix} 1&x_1&x_1^2&\dots&x_1^{n-1}\\ 1&x_2&x_2^2&\dots&x_2^{n-1}\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ 1&x_n&x_n^2&\dots&x_n^{n-1}\\ \end{vmatrix}=\prod\limits_{1\leq i<j\leq n}(x_j-x_i).$$ In short, we write $\det(x_i^{j-1})$. I became curious of a Hankel-type (not quite) formulation of these monomial entries as $\det(x_{i+j}^{j-1})$. A quick check reveal that the latter has no neat evaluation.

On the other hand, if we specialize $x_{i+j}$ to the numerical values $i+j$ then it appears that $$\det((i+j)^{j-1})=\prod_{k=1}^{n-1}k!\tag1$$

Question. Why is true?

Remark. The above evaluation equals the familiar $\det(i^{j-1})$ which OEIS lists as A000178 with many interpretations, including other determinants but (1) is not one of them.

Answer 1

Here is a slightly more general version: If $P_n(x)$ are some polynomials of degree $n-1$ with leading term $a_nx^{n-1}$ then $$\det(P_j(x_i))=\prod_{j=1}^n a_n\prod\limits_{1\leq i<j\leq n}(x_j-x_i).$$ The proof is a simple reduction to Vandermonde's determinant. The first column is all constant, and every column after that can be written as the vector $a_j(x_1^{j-1},\dots,x_n^{j-1})^{T}$ minus a linear combination of previous columns. Therefore a collection of elementary column operations bring our matrix to the form $(a_jx_i^{j-1})$.

Your determinant corresponds to the case $P_n(x)=(x+n)^{n-1}$.