On Wikipedia Here, the exponential generating function $$\sum_{n=k}^{\infty}{(-1)^{n-k}{n\brack k}\frac{z^n}{n!}}=\frac{1}{k!}(\log(1+z))^k$$ is given, where ${n\brack k}$ is the unsigned Stirling numbers of the first kind. I have done a literature search to see if I could find a similar but ordinary generating function for the unsigned Stirling numbers of the first kind, but I haven't found any.
Could it be that I am not doing a proper search, or no ordinary generating functions for the unsigned Stirling numbers of the first kind are known? Can someone refer me to some examples they might have seen?
I should mention that I have seen this one: $$\sum_{k=0}^{n}{{n\brack k}x^k}=x(x+1)(x+2)(x+3)\cdots(x+n-1),$$ but I am talking about a generating function of a similar kind where the upper summation index is infinity, just as in the case of the exponential generating function I quoted earlier.
This my guess at what you want. The “answer” is at the end in Wolfram. If not please clarify.
As usual: Please read carefully, I am subject to a mathematical dyslexia :)
Constructive comments welcome :)
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In order to “ground” the discussion below we start with definitions.
$x(x+1)\cdots(x+n-1)=\left(x\right)_{n}={\displaystyle \sum_{k=0}^{n}\left[\begin{array}{c} n\\ k \end{array}\right]x^{k}}$
$s(n,k)=\left(-1\right)^{n-k}\left[\begin{array}{c} n\\ k \end{array}\right]$
Notice that $\left(x\right)_{n}$ is the rising product. Which is opposite of Wikipedia; this is a persistent annoyance, not isolated to Wikipedia.
Let:
$G(n,k)={\displaystyle \sum_{n=0}^{\infty}{\displaystyle \sum_{k=0}^{n}\left[\begin{array}{c} n\\ k \end{array}\right]x^{k}}\cdot\frac{t^{n}}{n!}={\displaystyle \sum_{n=0}^{\infty}}\left(x\right)_{n}\cdot\frac{t^{n}}{n!}}$
Then we can use Wikipedia for
$G\left(n,k\right)=_{1}F_{0}\left(\begin{array}{c} x\\ - \end{array};t\right)=\left(1-t\right)^{-x}$ which rattles on (has an infinite number of terms) but does converge for t<1 (if your interested). We can eliminate the factorial by
$G\left(n,k\right)=_{2}F_{0}\left(\begin{array}{c} 1,x\\ - \end{array};t\right)$ which doesn't converge, but we can work term by term; if needed.
But
$H\left(n,k\right)={\displaystyle \sum_{n=0}^{\infty}}{\displaystyle \sum_{k=0}^{n}\left(-1\right)^{n-k}\cdot s(n,k)\cdot x^{k}}\cdot\frac{t^{n}}{n!}={\displaystyle \sum_{n=0}^{\infty}}\left(-x\right)_{n}\cdot\frac{\left(-t\right)^{n}}{n!}=_{1}F_{0}\left(\begin{array}{c} -x\\ - \end{array};-t\right)$
$=\left(1+t\right)^{x}={\displaystyle \sum_{n=0}^{x}\frac{x!}{\left(x-n\right)!\cdot n!}t^{n}}$
Does if $x\in N_{+}$
Remark 1. The last term looks strange until we untangle it.
${\displaystyle \sum_{n=0}^{x}\frac{x!}{\left(x-n\right)!\cdot n!}t^{n}}={\displaystyle \sum_{n=0}^{x}\frac{\Gamma\left(x+1\right)}{\Gamma\left(x-n+1\right)}\frac{t^{n}}{n!}}$
and apply the Gamma duality $\Gamma\left(z\right)=\frac{\pi}{\Gamma\left(1-z\right)\cdot sin(\pi\cdot z)}$ which for integer z is $\Gamma\left(z\right)=\frac{\left(-1\right)^{z}\cdot\pi}{\Gamma\left(1-z\right)}$
which yields
${\displaystyle \sum_{n=0}^{x}\frac{\Gamma\left(-x+n\right)}{\Gamma\left(-x\right)}\cdot\left(-1\right)^{n}\cdot\frac{t^{n}}{n!}=}{\displaystyle \sum_{n=0}^{x}\left(-x\right)_{n}\cdot\frac{\left(-t\right)^{n}}{n!}}$
Now when we remove the factorial we have
$_{2}\,F_{0}\left(\begin{array}{c} 1,-x\\ - \end{array};-t\right)$ does terminate and has representations at https://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/03/01/13/
Line 2 appears to work and is appealing; but that's due to a common programing bug. Don't "trust" computer algebra programs, find ways to double check them :)