After noticing that the determinant of an $n \times n$ matrix $A\_n$ with elements $a\_{i,j}=i^j$, $1 \le i \le n$, $1 \le j \le n$, is the superfactorial (product of the first $n$ factorials), I wanted to try other cases.


I noticed that with $a\_{i,j}=\binom{n+i-1}{j}$:


$$\lvert A\_n \rvert = \binom{2n-1}{n}$$


This one has been proved in [this answer](https://math.stackexchange.com/a/4455105/573047) using the paper [Binomial Determinants, Paths, and Hook Length Formulae](https://www.sciencedirect.com/science/article/pii/0001870885901215) by Gessel and Viennot.


After this I tried $a\_{i,j}=\left[{n+i-1\atop j}\right]$ (unsigned Stirling numbers of the first kind) and obtained:


$$\lvert A\_n \rvert = (n-1)!^{n} \tag{1}\label{1}$$


and with $a\_{i,j}={n+i-1 \brace j}$ (Stirling numbers of the second kind):


$$\lvert A\_n \rvert = n!^{n-1} \tag{2}\label{2}$$


Any hint for proving $(1)$ and $(2)$?


@Peter Taylor found the following generalizations:


With $a\_{i,j}=\left[{k+i-1\atop j}\right]$ (unsigned Stirling numbers of the first kind):


$$\lvert A\_n \rvert = (k-1)!^{n} \tag{3}\label{3}$$


and with $a\_{i,j}={k+i-1 \brace j}$ (Stirling numbers of the second kind):


$$\lvert A\_n \rvert = n!^{k-1} \tag{4}\label{4}$$



Determinant of matrix with Stirling numbers as elements
If we take a general sequence of (unsigned) Comtet numbers of the first kind [1, 2] $$c(n, k) = e\_{n-k}(\xi\_1, \ldots, \xi\_n)$$ then the $n \times n$ submatrix with a row offset of $k$ has determinant $$\det\_{0 \le i, j < n} \Big( e\_{k+i-j}(\xi\_1, \ldots, \xi\_{k+i}) \Big) = (\xi\_1 \cdots \xi\_k)^n$$


Similarly, if we take a general sequence of Comtet numbers of the second kind $$C(n, k) = h\_{n-k}(\xi\_1, \ldots, \xi\_{k+1})$$ then the $n \times n$ submatrix with a row offset of $k$ has determinant $$\det\_{0 \le i,j < n} \Big( h\_{k+i-j}(\xi\_1, \ldots, \xi\_{j+1}) \Big) = (\xi\_1 \cdots \xi\_n)^k$$


Note in particular that for your question (the Stirling numbers respectively of the first and second kinds, omitting the row $n=0$ and column $k=0$) we can take $\xi\_i = i$.


A more natural generalisation of both the binomial and the Stirling examples you give is to take a row offset of $k$ and a column offset of $1$, but the resulting LU factorisations are far uglier.


### Proofs


In a comment, [Max Muller](https://mathoverflow.net/users/93724/max-muller) gave a reference to C. Krattenthaler, *[Advanced determinant calculus: A complement](https://doi.org/10.1016/j.laa.2005.06.042)*, Lin. Alg. and its Applications, 411 (2005), 68-166. Theorems 44 and 45 are very much in the same area, certainly generalise the result for Comtet numbers of the first kind. I'm no longer confident that they also cover Comtet numbers of the second kind, but I haven't wrestled with them enough to say that they definitely don't. Regardless, the paper which Krattenthaler references for these results is described as "in press" and, 17 years later, doesn't appear yet to have been published, so I give proofs for the more restricted claims made above.


**Theorem:** $\det\_{0 \le i, j < n} \Big( e\_{k+i-j}(\xi\_1, \ldots, \xi\_{k+i}) \Big) = (\xi\_1 \cdots \xi\_k)^n$


Proof: $\Big( e\_{k+i-j}(\xi\_1, \ldots, \xi\_{k+i}) \Big)\_{0 \le i, j < n} = L\_e U\_e$ where $$\begin{eqnarray\*}L\_e &=& \Big( e\_{i-j}(\xi\_{k+1}, \ldots, \xi\_{k+i}) \Big)\_{0 \le i, j < n} \\
U\_e &=& \Big( e\_{k+i-j}(\xi\_1, \ldots, \xi\_k) \Big)\_{0 \le i, j < n}
\end{eqnarray\*}$$


To verify the factorisation observe that $$(L\_e U\_e)\_{i,j} = \sum\_{a=0}^{n-1} e\_{i-a}(\xi\_{k+1}, \ldots, \xi\_{k+i}) e\_{k+a-j}(\xi\_1, \ldots, \xi\_k) = e\_{k+i-j} (\xi\_1, \ldots, \xi\_{k+i})$$


Trivially, $\det L\_e = 1$ and $\det U\_e = e\_k(\xi\_1, \ldots, \xi\_k)^n = (\xi\_1 \cdots \xi\_k)^n$.  

$\blacksquare$


**Theorem:** $\det\_{0 \le i,j < n} \Big( h\_{k+i-j}(\xi\_1, \ldots, \xi\_{j+1}) \Big) = (\xi\_1 \cdots \xi\_n)^k$


Again we use an LU-factorisation as $L\_h U\_h$ where $$\begin{eqnarray\*}L\_h &=& \Big( h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) \Big)\_{0 \le i, j < n} \\
U\_h &=& \Big( h\_{k+i-j}(\xi\_{i+1}, \ldots \xi\_{j+1}) \Big)\_{0 \le i, j < n}
\end{eqnarray\*}$$


I don't find that the verification yields to proof by inspection this time because of the overlap in variables between the two halves of each term:


$$(L\_h U\_h)\_{i,j} = \sum\_{a=0}^{n-1} h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-j}(\xi\_{a+1}, \ldots \xi\_{j+1})$$


So let $P(i,j,k, \vec{\xi}) = \sum\_{a=0}^{j} h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-j}(\xi\_{a+1}, \ldots \xi\_{j+1})$ and we aim to prove by induction on $j$ that $P(i,j,k,\xi) = h\_{i+k-j}(\xi\_1, \ldots, \xi\_{j+1})$.


Base case, $j=0$: $P(i,0,k, \vec{\xi}) = h\_{i}(\xi\_1) h\_{k-j}(\xi\_{1}) = h\_{i+k-j}(\xi\_i)$


Inductive step, $j > 0$: \begin{eqnarray\*}
P(i,j,k, \vec{\xi}) &=& \sum\_{a=0}^{j} h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-j}(\xi\_{a+1}, \ldots \xi\_{j+1}) \\
&=& h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) h\_{k}(\xi\_{j+1}) + \sum\_{a=0}^{j-1} h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-j}(\xi\_{a+1}, \ldots \xi\_{j+1}) \\
&=& \xi\_{j+1}^k h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) + \sum\_{a=0}^{j-1} \sum\_{b=0}^{k+a-j} \xi\_{j+1}^b h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-b-j}(\xi\_{a+1}, \ldots \xi\_{j}) \\
&=& \xi\_{j+1}^k h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) + \sum\_{b=0}^{k-1} \xi\_{j+1}^b \sum\_{a=0}^{j-1} h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-b-j}(\xi\_{a+1}, \ldots \xi\_{j}) \\
&=& \xi\_{j+1}^k h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) + \sum\_{b=0}^{k-1} \xi\_{j+1}^b P(i,j-1,k-b-1, \vec{\xi}) \\
&=& \xi\_{j+1}^k h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) + \sum\_{b=0}^{k-1} \xi\_{j+1}^b h\_{i+k-b-j}(\xi\_1, \ldots, \xi\_j) \\
&=& h\_{i+k-j}(\xi\_1, \ldots, \xi\_{j+1})
\end{eqnarray\*}


By design, we have $(L\_h U\_h)\_{i,j} = P(i,j,k, \vec{\xi})$, with the difference in upper limit of the sum being trivial due to the term $h\_{k+a-j}(\xi\_{a+1}, \ldots \xi\_{j+1})$. Then $\det L\_h = 1$ and $\det U\_h = h\_k(\xi\_1) \cdots h\_k(\xi\_n) = (\xi\_1 \cdots \xi\_n)^k$.  

$\blacksquare$


[1] Louis Comtet, [Nombres de Stirling généraux et fonctions symétriques](https://gallica.bnf.fr/ark:/12148/bpt6k5619091b/f23.item), C. R. Acad. Sc. Paris, t. 275 (1972), Sér. A 747–750.  

[2] Hopkins, Speyer, Taylor, Zaimi, [Matrices of combinatorial sequences that are inverse in two ways](https://mathoverflow.net/q/418266/46140) , MO question 418266 and answers


