I would like to sum the following series:
$$\sum_{n=0}^\infty \frac{u^n}{n!}\sum_{k=1}^{n+1}(-1)^{k-1}(k-1)! S_{n+1,k}p^k,$$
where $S_{n,k}$ are Stirling numbers of the second kind, $0<p<1, \ 0<u<1.$ We can set $p=u,$ if that makes things easier.
I tried to attack this with generating functions, and came up with something that seems certainly wrong, but it seems like a clean closed form solution may exist.
We seek to simplify
$$\sum_{n\ge 0} \frac{u^n}{n!} \sum_{k=1}^{n+1} (-1)^{k-1} (k-1)! {n+1\brace k} p^k.$$
Using formal power series we have
$${n+1\brace k} = (n+1)! [z^{n+1}] \frac{(\exp(z)-1)^k}{k!}.$$
We obtain for the sum
$$\sum_{n\ge 0} \frac{u^n}{n!} (n+1)! [z^{n+1}] \sum_{k=1}^{n+1} (-1)^{k-1} \frac{1}{k} (\exp(z)-1)^k p^k \\ = \sum_{n\ge 0} u^n (n+1) [z^{n+1}] \log\left(1+p(\exp(z)-1)\right) \\ = \left.\log\left(1+p(\exp(z)-1)\right)'\right|_{z=u}.$$
This yields $$\bbox[5px,border:2px solid #00A000]{ \frac{p\exp(u)}{1+p(\exp(u)-1)}.}$$