While playing around with the MO question https://mathoverflow.net/questions/427934/determinant-with-factorials-is-not-0 about a determinant of the Hankel matrix of entries $(i+j-2)!$, having the value $\prod_{k=0}^{n-1}k!^2$, I stumbled on the following. 

A permutation $\pi\in\mathfrak{S}_n$ is called a [*derangement*][2] if it has no fixed points. Let $d_n$ be the number of derangement permutations in $\mathfrak{S}_n$ which may be presented by the formula
$$d_n=n!\sum_{k=0}^n\frac{(-1)^k}{k!}.$$

>**QUESTION.** It appears that we have $\det((i+j-2)!)=\det(d_{i+j-2})$. Why? Why not?

[1]: https://mathoverflow.net/questions/427934/determinant-with-factorials-is-not-0
[2]: https://en.wikipedia.org/wiki/Derangement