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 gave a reference to C. Krattenthaler, *Advanced determinant calculus: A complement*, 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, 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 , MO question 418266 and answers