In my earlier (soft) MO post, an elementary response 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 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*.