In my earlier (soft) [MO post](https://mathoverflow.net/questions/415230/on-a-variation-of-the-vandermonde-matrix), an elementary [response](https://mathoverflow.net/questions/415230/on-a-variation-of-the-vandermonde-matrix#comment1064945_415230) was given by Ofir Gorodetsky in regard to the determinant of the symbolic counterpart to the numerical matrix $\mathbf{M}\_n=(i^j-j^i)\_{i,j}^{1,n}$.


I'm now looking at yet another variation. Let $\mathbf{A}\_n=(j^j-i^j)\_{i,j}^{1,n}$ be an $n\times n$-matrix. Denote the (signed) [Stirling numbers of the first kind](https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind) by $s(n,k)$. For clarity, here are some contrasting examples:
$$\mathbf{M}\_3=\begin{pmatrix} 0&-1&-2 \\ 1&0&-1 \\2&1&0 \end{pmatrix}
\qquad \text{and} \qquad \mathbf{A}\_3=\begin{pmatrix}0&3&26 \\-1&0&19 \\ -2&-5&0 \end{pmatrix}.$$
I would like to ask:



> 
> **QUESTION.** Is this true? With the convention that $0^0=1$, we have
> $$\det\mathbf{A}\_n=\prod\_{j=1}^{n-1}j!\cdot\sum\_{k=0}^n\,s(n+1,k+1)\cdot k^k.$$
> 
> 
> 


**Remark.** If it helps, since the factor $\prod\_{j=1}^{n-1}j!$ is exactly the determinant of the Vandermonde matrix, $\mathbf{V}\_n=(i^{j-1})\_{i,j}^{1,n}$, we can say
$$\frac{\det\mathbf{A}\_n}{\det \mathbf{V}\_n}=\sum\_{k=0}^n\,s(n+1,k+1)\cdot k^k.$$ This formalism goes in the "spirit" of the (specialization) $s\_{\lambda}(1,\cdots,1)$ of the [*Schur polynomials*](https://en.wikipedia.org/wiki/Schur_polynomial).



Yet, another numerical variant of the Vandermonde matrix
This is true and this is not about numbers $j^j$ at all. Consider a more genral matrix $(c\_j-i^j)\_{1\leqslant i,j\leqslant n}$. Denote $f\_j(x)=c\_j-x^j$, and find the numbers $\beta$, $\alpha\_1,\ldots,\alpha\_{n-1}$ such that $f\_n(x)+\sum\_{j=1}^{n-1}\alpha\_j f\_j(x)=\beta$ for all $x\in \{1,2,\ldots,n\}$. Then the matrix $(f\_j(i))\_{i,j}$ has the same determinant as if we replace the column with $f\_n$ by a column of $\beta$'s, that is seen from operations with columns. Now by operations with columns we may replace each $f\_j$ to $-x^j$, so we get almost a Vandermonde matrix (up to $n-1$ changes of signs in the columns and the cyclic shift, and $\beta$'s instead of 1's), so its determinant equals to the determinant of Vandermonde matrix for the numbers $1,\ldots,n$ times $(-1)^{n-1}(-1)^{n-1}\beta=\beta$. Thus, it appears to find $\beta$. For this, write
$$
f\_n+\sum\_{j=1}^{n-1}\alpha\_jf\_j(x)=\beta-(x-1)(x-2)\ldots(x-n)=
\beta-\sum\_{k=0}^n s(n+1,k+1)x^k.
$$
Equalising the coefficients of $x^k$ we get $\alpha\_k=s(n+1,k+1)$ for all $k=1,\ldots,n$ (where $\alpha\_n=1$ by agreement), thus $\beta-s(n+1,1)=\sum\_{k=1}^n c\_k s(n+1,k+1)$, $\beta=\sum\_{k=0}^n c\_k s(n+1,k+1)$, where $c\_0=1$ by agreement.


