While playing around with this MO question 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 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?

While playing around with the MO question 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 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?

While playing around with this MO question about a determinant of the Hankel matrix of entries $(i+j-2)!$, I stumbled on the following.

A permutation $\pi\in\mathfrak{S}_n$ is called a derangement 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?

While playing around with this MO question 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 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?