Finite sums involving Stirling number of first kind I would like simplify the following doble sum

$$ \sum_{k=m}^n\,s(n,k)\,x^k\sum_{s=m}^k\,(-1)^{k+s}\,s(k,s)\begin{pmatrix}s\\m\end{pmatrix}\,y^{s-m}$$

with $s(n,k)$ the Stirling numbers of first kind. I've be able to invert the order of summation
$$ \sum_{s=m}^n\begin{pmatrix}s\\m\end{pmatrix}\,y^{s-m}\sum_{k=s}^n\,(-1)^{k+s}\,s(n,k)\,s(k,s)\,x^k$$
but the difficulty seems to be the same.
Any help will be welcomed.

Let me change a bit the symbols and write
$$
\eqalign{
  & M_{\,n,\,m} (x,y) = \sum\limits_{k = m}^n {
 \left[ \matrix{  n \cr   k \cr}  \right]x^{\,k} \sum\limits_{j = m}^k {
 \left( { - 1} \right)^{\,k + j} \left[ \matrix{  k \cr   j \cr}  \right]\left( \matrix{  j \cr   m \cr}  \right)y^{\,j - m} } }  =   \cr 
  &  = \sum\limits_{\left( {m\, \le } \right)\,k\,\left( { \le \,n} \right)} {
 \sum\limits_{\left( {m\, \le } \right)\,j\,\left( { \le \,k} \right)} {
 \left( { - 1} \right)^{\,k + j} \left[ \matrix{  n \cr   k \cr}  \right]\left[ \matrix{  k \cr   j \cr}  \right]
 \left( \matrix{  j \cr   m \cr}  \right)x^{\,k} y^{\,j - m} } }  =   \cr 
  &  = \sum\limits_{\left( {m\, \le } \right)\,k\,\left( { \le \,n} \right)} {
 \sum\limits_{\left( {m\, \le } \right)\,j\,\left( { \le \,k} \right)} {
 \left[ \matrix{  n \cr   k \cr}  \right]x^{\,k} \left( { - 1} \right)^{\,k} \left[ \matrix{  k \cr  j \cr}  \right]
 \left( { - 1} \right)^{\,j} y^{\,j} \left( \matrix{  j \cr   m \cr}  \right)y^{\, - m} } }  \tag{1} \cr} 
$$
where the bounds of the sums are put in brackets to show that they are actually implicit
in the binomial and Stirling 1st kind.
Now we have that
$$
\eqalign{
  & x^{\,\overline {\,n\,} }  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
 \left[ \matrix{  n \cr   k \cr}  \right]x^{\,k} } \quad  \Rightarrow   \cr  &  \Rightarrow \quad \left( {x + y} \right)^{\,\overline {\,k\,} }
  = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {
 \left[ \matrix{  k \cr   j \cr}  \right]\left( {x + y} \right)^{\,j} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {
 \sum\limits_{\left( {0\, \le } \right)\,m\,\left( { \le \,j} \right)} {
 \left[ \matrix{  k \cr   j \cr}  \right]\left( \matrix{  j \cr   m \cr}  \right)x^{\,m} y^{\,j - m} } } \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( { - x - y} \right)^{\,\overline {\,k\,} }
  = \left( { - 1} \right)^{\,k} \left( {x + y} \right)^{\,\underline {\,k\,} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,m\,\left( { \le \,k} \right)} {
 \left( {\sum\limits_{\left( {m\, \le } \right)\,j\,\left( { \le \,k} \right)} {
 \left[ \matrix{  k \cr  j \cr}  \right]\left( \matrix{  j \cr   m \cr}  \right)\left( { - 1} \right)^{\,j} x^{\,m} y^{\,j - m} } } \right)}
  \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \sum\limits_{\left( {m\, \le } \right)\,j\,\left( { \le \,k} \right)} {
 \left[ \matrix{  k \cr   j \cr}  \right]\left( \matrix{  j \cr   m \cr}  \right)\left( { - 1} \right)^{\,j} y^{\,j - m} }
  = \left[ {x^{\,m} } \right]\left( { - x - y} \right)^{\,\overline {\,k\,} }  \cr} 
$$
Therefore
$$
\eqalign{
  & M_{\,n,\,m} (x,y) =   \cr 
  &  = \sum\limits_{\left( {m\, \le } \right)\,k\,\left( { \le \,n} \right)} {
 \left[ \matrix{  n \cr   k \cr}  \right]x^{\,k} \left( {\left[ {x^{\,m} } \right]\left( {x + y} \right)^{\,\underline {\,k\,} } } \right)}  =   \cr 
  &  = {1 \over {m!}}\sum\limits_{\left( {m\, \le } \right)\,k\,\left( { \le \,n} \right)} {
 \left[ \matrix{  n \cr   k \cr}  \right]x^{\,k} {{d^{\,m} } \over {dy^{\,m} }}y^{\,\underline {\,k\,} } }  \cr \tag{2}} 
$$
where $\left[ {x^{\,m} } \right]$ stands for "the coefficient of $x^m$ (in the following expression)".
