Is there a way to prove the following identity
\begin{equation} \sum_{l = 1}^{k} \left( \frac{(-s)^l}{l!} \sum_{n_1 + n_2 + \ldots n_l = k} \frac{1}{n_1n_2 \ldots n_k} \right)= (-1)^k {s \choose k} \, , \end{equation} where $k$ and $n_i$'s are positive integer and $s$ is a real number. I have checked this identity on Mathematica for several $k$'s.
Consider the equality $(1-x)^s=\exp(s\log(1-x))$. Since $$\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}\quad\text{and}\quad\log(1-x)=-\sum_{m=1}^\infty\frac{x^m}{m},$$
we have $$ \begin{aligned} \exp(s\log(1-x))&=\exp\left(-s\sum_{m=1}^\infty\frac{x^m}{m}\right)\\ &=\sum_{n=0}^\infty\frac{1}{n!}\left(-s\sum_{m=1}^\infty\frac{x^m}{m}\right)^n\\ &=\sum_{n=0}^\infty\frac{(-s)^n}{n!}\left(\sum_{m=1}^\infty\frac{x^m}{m}\right)^n. \end{aligned}$$
Thus the coefficient of $x^k(k>0)$ of $\exp(s\log(1-x))$ is $$\sum_{l = 1}^{k} \left( \frac{(-s)^l}{l!} \sum_{n_1 + n_2 + \cdots +n_l = k} \frac{1}{n_1n_2 \ldots n_k} \right).$$
On the other side, the coefficient of $x^k$ of $(1-x)^s$ is $(-1)^k {s \choose k}$. Now we get the desired result.