While playing around with the MO question [Determinant with factorials is not 0?](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*](https://en.wikipedia.org/wiki/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
$$d\_n=n!\sum\_{k=0}^n\frac{(-1)^k}{k!}.$$



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> **QUESTION.** It appears that we have $\det((i+j-2)!)=\det(d\_{i+j-2})$. Why? Why not?
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A coincidence or a fact: determinants of two matrices
Orangeskid's guess is correct: a more general fact holds that the binomial transform preserves Hankel determinants.


For a matrix $(a\_{ij})$ (it is convenient to enumerate rows and columns from 0,not from 1) denote $$b\_{ij}=\sum\_{k, s}{i\choose k}{j\choose s}a\_{ks}.$$
This matrix transform corresponds to a left and right multiplicaton by unitriangular matrices, thus it preserves determinant. Now if $a\_{ij}=f(i+j)$ is a Hankel matrix, then $$b\_{ij}=\sum\_t f(t)\sum\_{k+s=t} {i\choose k}{j\choose s}=\sum\_tf(t){i+j\choose t}$$
is a Hankel matrix corresponding to the binomial transform of $f$.


It remains to recall that $n! =\sum\_k {n\choose k} d\_k$ (combinatorially ${n\choose k} d\_k$ counts the number of permutations of $\{1,\ldots, n\}$ with exactly $n-k$ fixed points, thus this formula), that means that the sequence of factorials is the binomial transform of the sequence of dearrangements.


