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](https://oeis.org/A000178) with many interpretations, including other determinants but (1) is not one of them.



"Almost Hankelized" numerical Vandermonde
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 $P\_n(x)=(x+n)^{n-1}$.


