Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $n \brace k$ be a Stirling number of the second kind.
Let
$$ f(n,m,i) = (-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace j} $$
I conjecture that
$$ f(n,m,i) = \sum\limits_{j=i-1}^{m} \binom{m+1}{j} f(n-1, j, i), \\ f(0,m,i) = [i = (m+1)]. $$
Here square bracket denotes Iverson bracket.
Is there a way to prove it? Is there a way to find first formula if the last one is known?
Noticing that $j^n = n![x^n]\ e^{jx}$ and $\left\{m+1\atop j\right\}=(m+1)![y^{m+1}]\ \frac{(e^y-1)^j}{j!}$, and using e.g.f. for Stirling numbers of first kind, we have $$f(n,m,i) = (-1)^{m-i+1} \frac{(m+1)! n!}{i!} [x^ny^{m+1}] \log(1+e^x(e^y-1))^i.$$
Then \begin{split} S&:=\sum_j \binom{m+1}j f(n-1,j,i) \\ & = \sum_j \binom{m+1}j (-1)^{j-i+1} \frac{(j+1)! (n-1)!}{i!} [x^{n-1}y^{j+1}] \log(1+e^x(e^y-1))^i \\ & = (-1)^{m-i} \frac{(n-1)!(m+1)!}{i!} [x^{n-1}y^{m+1}]\ e^{-y} \frac{\partial}{\partial y}\log(1+e^x(e^y-1))^i \end{split} and thus \begin{split} \sum_{j\leq m} & \binom{m+1}j f(n-1,j,i) \\ & = S - f(n-1,m+1,i) \\ & = S - (-1)^{m-i} \frac{(m+2)! (n-1)!}{i!} [x^{n-1}y^{m+2}] \log(1+e^x(e^y-1))^i \\ & = S - (-1)^{m-i} \frac{(m+1)! (n-1)!}{i!} [x^{n-1}y^{m+1}] \frac{\partial}{\partial y}\log(1+e^x(e^y-1))^i \\ & = (-1)^{m-i} \frac{(m+1)! (n-1)!}{i!} [x^{n-1}y^{m+1}] (e^{-y}-1)\frac{\partial}{\partial y}\log(1+e^x(e^y-1))^i \\ & = (-1)^{m-i+1} \frac{(m+1)! (n-1)!}{i!} [x^{n-1}y^{m+1}] \frac{\partial}{\partial x}\log(1+e^x(e^y-1))^i \\ & = (-1)^{m-i+1} \frac{(m+1)! n!}{i!} [x^n y^{m+1}] \log(1+e^x(e^y-1))^i \\ & = f(n,m,i), \end{split} where we used the observation that $$(e^{-y}-1)\frac{\partial}{\partial y}\log(1+e^x(e^y-1))^i = - \frac{\partial}{\partial x}\log(1+e^x(e^y-1))^i.$$ QED
