# Identity of binomial series with factorial: $\sum_{k = 0}^{p} \binom{p}{k} \cdot k! \cdot x^k$

I'm looking for a simple identity for the formula:

$$\sum_{k = 0}^{p} \binom{p}{k} \cdot k! \cdot x^k$$

In words, I have $p$ "players" who can choose to play or not (every player is represented by a unique id). Those who chose to play are lined up in all possible orders. Then every playing player picks an element out of $x$ possibilities (with repetition allowed), in addition to his id.

How many sequences can we get? Is there a simple solution for this series? If not, what is the closest upper limit you can think of?

I haven't touched combinatorics for a long time, so there could be a simple identity that I'm missing...

There is no "simple" solution: $$\sum_{k=0}^{p} \binom{p}{k} \cdot k! \cdot x^k = \sum_{k=0}^{p} \frac{p!}{(p-k)!} \cdot x^k \underset{k' = p-k}{=} \sum_{k=0}^{p} \frac{p!}{k!} \cdot x^{p-k} = p!\cdot x^p \underbrace{\sum_{k=0}^{p} \frac{\left(\frac{1}{x}\right)^k}{k!}}_ {\underset{p \to \inf}{\rightarrow} e^{1/x}} \leq p! \cdot x^p \cdot e^{{}^\frac{1}{x}}$$
You can rewrite the formula as $$\sum_{k = 0}^{p} \binom{p}{k} \cdot k! \cdot x^k = \sum_{k = 0}^{p}\frac{p!}{(p-k)!}x^k = 1+px+p(p-1)x^2+\ldots+p!x^p.$$