A Vandermonde like determinant with exponentials Let $n\geq m$ be non negative integers, and consider a list of $(n+m+1)$ distinct numbers (complex or real). I am interested in getting a closed form formula for the following determinant: $\det\left[\sum_{k=i}^{n+i}s_k^{m+1-j}e^{-s_k}\left[\prod_{l=i,l\neq k}^{n+i}(s_k-s_l)\right]^{-1}\right]_{i,j=1\ldots m+1}$, i.e.
$$
\det\left(\begin{array}[ccc]
 \,\sum_{k=1}^{n+1}s_k^me^{-s_k}\left[\prod_{l=1,l\neq k}^{n+1}(s_k-s_l)\right]^{-1}& \ldots &\sum_{k=1}^{n+1}e^{-s_k}\left[\prod_{l=1,l\neq k}^{n+1}(s_k-s_l)\right]^{-1}  \\ 
\vdots & \ddots & \vdots \\ 
\sum_{k=m+1}^{n+m+1}s_k^me^{-s_k}\left[\prod_{l=m+1,l\neq k}^{n+m+1}(s_k-s_l)\right]^{-1}& \ldots &\sum_{k=m+1}^{n+m+1}e^{-s_k}\left[\prod_{l=m+1,l\neq k}^{n+m+1}(s_k-s_l)\right]^{-1}
\end{array}\right)
$$.
This determinant can be rewritten as the determinant of an alternant matrix $\left(F_{j}(S_i)\right)_{i,j=1\ldots m+1}$, for suitable multivariate functions $F_j:\mathbb{C}^{n+1}\rightarrow\mathbb{C}$.
If the exponential are not there, then this determinant should some king of Schur polynomial. I am wondering if this determinant can be interpreted in terms of representation theory.
A related post : A class of matrix determinants between Wronskians and Vandermondes
 A: After slight renumbering $(m,n)\mapsto (m,n)-1$ and horizontal reflection $j\mapsto m+1-j$, let's call the $m\times m$ matrix in the OP's determinant
$$
A=\left[\sum_{k=i}^{i+n-1}s_k^{j-1}e^{-s_k}\prod_{l=i,l\neq k}^{i+n-1}(s_k-s_l)^{-1}\right]_{\,i,j=1}^{\,m},\tag{1}
$$
and define $r=m+n-1$. Note that this answer also holds for $n<m$.
$A$ can be factorized similar to this posting using the $r \times m$ Vandermonde matrix
$$
V_{r,m}=\left[ s_\rho^{\mu-1}\right]_{\rho,\mu=1}^{r,m},\tag{2}
$$
the diagonal matrix
$$
E_r = \mathrm{diag}\,[e^{-s_\rho}]_{\rho=1}^r\tag{3}
$$
and the matrix of inverse derivatives
$$
D_{m,r}^{(o)}=\left[ \frac{1}{P_{\mu,o}'(s_\rho)}
 \begin{cases} 
  1 & \mu \leq \rho < m+\mu \\
  0 & \text{else}
 \end{cases}
\right]_{\,\mu,\rho=1}^{\,m,r},\tag{4}
$$
of the polynomials
$$
P_{l,o}(s)=\prod_{\rho=l}^{l+o-1}(s-s_\rho)\tag{5}
$$
according to
$$
A = D_{m,r}^{(m)} E_{r} \, V_{r,m}.\tag{6}
$$
Note that $D$ is not diagonal in this case, e.g., for $m=3$, $n=3$, where $r=5$:
$$
D_{3,5}^{(3)}=
\left(
\begin{array}{ccccc}
 \frac{1}{\left(s_1-s_2\right) \left(s_1-s_3\right)} &
   \frac{1}{\left(s_2-s_1\right) \left(s_2-s_3\right)} &
   \frac{1}{\left(s_3-s_1\right) \left(s_3-s_2\right)} & 0 & 0 \\
 0 & \frac{1}{\left(s_2-s_3\right) \left(s_2-s_4\right)} &
   \frac{1}{\left(s_3-s_2\right) \left(s_3-s_4\right)} &
   \frac{1}{\left(s_4-s_2\right) \left(s_4-s_3\right)} & 0 \\
 0 & 0 & \frac{1}{\left(s_3-s_4\right) \left(s_3-s_5\right)} &
   \frac{1}{\left(s_4-s_3\right) \left(s_4-s_5\right)} &
   \frac{1}{\left(s_5-s_3\right) \left(s_5-s_4\right)} \\
\end{array}
\right).\tag{7}
$$
As the matrices $D$ and $V$ are not square, the determinant of $A$ does not factorize in a simple way. However, one could try to enlarge both matrices to $r \times r$ by inserting appropriate matrix elements $O(\epsilon)$, factorize and let $\epsilon\to 0$. Alternatively, one can naturally enlarge $D$ and $V$ and define the $r \times r$ matrix
$$
B=D_{r,r}^{(m)} E_{r} \, V_{r,r},\tag{8}
$$
where $\det B$ factorizes. The matrix $A$ then is the upper left $m \times m$ submatrix of $B$.
