Closed form of $\sum_{k=1}^{n} s(n, k) k x^{k}=?$

Question

My question is

$$\sum_{k=1}^{n} s(n, k) k x^{k}=?$$

$s(n, k)$ are Stirling numbers of the first kind. Here is a reference for second kind of stirling numbers.

Formula for $\sum_{k=0}^n S(n,k) k$, where $S(n,k)$ is a Stirling number of the second kind? thanks for your answers.

Is there any reference about it?

Answer 1

In terms of special functions $$S_n=\sum_{k=1}^n s(n,k)\,k\,x^k=x\left(\sum_{k=0}^n s(n,k)\,x^k\right)'=x \left(x^{(n)}\right)'$$ where appears the factorial power or falling factorial function. $$S_n=x \,x^{(n)}\,(\psi (x+1)-\psi (x+1-n))=x \,x^{(n)} \left(H_x-H_{x-n}\right)$$

We have $$x^{(1)}=x \qquad \text{and} \qquad \frac{x^{(n+1)}}{x^{(n)}}=x-n$$

Answer 2

$$\sum_{k=1}^n s(n,k)kx^k=x\frac{d}{dx}\sum_{k=0}^n s(n,k)x^k=x\frac{d}{dx}\prod_{k=0}^{n-1}(x-k)=\left(\sum_{k=0}^{n-1}\frac{x}{x-k}\right)\prod_{k=0}^{n-1}(x-k)$$ according to one of the definitions. The last expression doesn't seem to simplify.