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


$$
\sum\_{n = 0}^{p} \binom{p}{n} \cdot n! \cdot x^n
$$


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...



Identity of binomial series with factorial.
Darij's comment... 


Tthe truncated exponential series:
$$
e\_p(z) = \sum\_{n=0}^p\frac{z^n}{n!}
$$
Your sum is
$$
p!x^pe\_p(1/x)
$$


