There is a set $S=\{x_1, x_2, ..., x_N\}.$
I'm trying to approximate this: $$p(S)=\sum_{r\subset S}|r|!\prod_{x\in r}x$$
I know that:
$$\sum_{r\subset S}\prod_{x\in r}x=\prod_{x\in S}(1+x)$$
I was wondering if there is a way to approximate $p(S)$ with something.
An idea: Change $x$ in $\prod\limits_{x\in S}(1+x)$ to $a(x)x$ so that:
$$\prod_{x\in S}(1+a(x)x)\sim\sum_{r\subset S}|r|!\prod_{x\in r}x$$
Stirling's approximation
There is: $$n! \sim (2\pi n)^\frac{1}{2}(\frac{n}{e})^n$$ which for my problem $n^n$ is troubling and I can't fiure out a way for $\prod_{x\in S}(1+a(x)x)$ to make $n^n$. It could also go up to a power of e: $$n! \sim e^{log(2\pi)/2-n+nlog(n)}$$ But again, can't figure out to handle $nlog(n)$.
We can find an easy upper bound for $p(S)$ in the function $\prod_{x\in S}(1+x^2)$, as $$\prod_{x\in S}(1+x^2)=\sum_{r\subset S}\prod_{x\in r}x^2=\sum_{r\subset S}\bigg(\prod_{x\in r}x\bigg)^2=\sum_{r\subset S}\bigg(\prod_{x\in r}x\bigg)\prod_{x\in r}x\geq\sum_{r\subset S}|r|!\prod_{x\in r}x=p(S)$$
(the inequality holding as all $x$'s are positive integers)
For a lower bound, you could use the product $\prod_{x\in S}(1+x)$, as you have mentioned.
Correcting what I wrote in a comment, you could also use $\prod_{x\in S}(2+x)-R(S)$ where $R(S)$ is a remainder term.
$$p(S)=\sum_{r\subset S}|r|!\prod_{x\in r}x=\\\sum_{r\subset S,|r|=0}0!\prod_{x\in r}x+\sum_{r\subset S,|r|=1}1!\prod_{x\in r}x+\sum_{r\subset S,|r|=2}2!\prod_{x\in r}x+\sum_{r\subset S,|r|=3}3!\prod_{x\in r}x+\sum_{r\subset S,|r|>3}|r|!\prod_{x\in r}x=\\ 1+\sum_{x\in S}x+2\sum_{x,y\in S\\x\neq y}xy+6\sum_{x,y,z\in S\\x\neq y\neq z}xyz+\sum_{r\subset S,|r|>3}|r|!\prod_{x\in r}x\geq\\ 1+\sum_{x\in S}x+2\sum_{x,y\in S\\x\neq y}xy+6\sum_{x,y,z\in S\\x\neq y\neq z}xyz+\sum_{r\subset S,|r|>3}2^{|r|}\prod_{x\in r}x\geq\\ 1+\sum_{x\in S}x+2\sum_{x,y\in S\\x\neq y}xy+6\sum_{x,y,z\in S\\x\neq y\neq z}xyz+\sum_{r\subset S\\|r|>3}\sum_{n\in 2^{|r|}}\prod_{x\in n}x=\\1+\sum_{x\in S}x+2\sum_{x,y\in S\\x\neq y}xy+6\sum_{x,y,z\in S\\x\neq y\neq z}xyz+\sum_{r\subset S\\|r|>3}\sum_{n\subset r}\prod_{x\in n}x=\\ 1+\sum_{x\in S}x+2\sum_{x,y\in S\\x\neq y}xy+6\sum_{x,y,z\in S\\x\neq y\neq z}xyz+\sum_{r\subset S\\|r|>3}\prod_{x\in r}(1+x)=\\\bigg(1+\sum_{x\in S}x+2\sum_{x,y\in S\\x\neq y}xy+6\sum_{x,y,z\in S\\x\neq y\neq z}xyz\bigg)+\sum_{r\subset S}\prod_{x\in r}(1+x)-\bigg(1+\sum_{x\in S}(1+x)+\sum_{x,y\in S\\ x\neq y}(1+x)(1+y)+\sum_{x,y,z\in S\\x\neq y\neq z}(1+x)(1+y)(1+z)\bigg)=\\\sum_{r\subset S}\prod_{x\in r}(1+x)-|S|-\binom{|S|}{2}-\binom{|S|}{3}-\frac{|S|(|S|-1)}{2}\sum_{x\in S}x-(|S|-3)\sum_{x,y\in S\\x\neq y}xy+\\5\sum_{x,y,z\in S\\x\neq y\neq z}xyz=\\\sum_{r\subset S}\prod_{x\in r}(1+x)-R(S)=\sum_{t\subset S+1}\prod_{y\in t}y-R(S)=\prod_{y\in S+1}(1+y)-R(S)=\\\prod_{x\in S}(2+x)-R(S)$$
(I have used expressions like $S+1$ above to denote the set consisting of elements of $S$ each incremented by $1$)
The upper bound represents setting $a(x)=x$ and the lower bound, setting $a(x)=1+\frac{1}{x}$, both, of course, diverge from $p(S)$ vastly for large sizes of $S$.
I can see no way of finding a function $a(x)$ that would satisfy the constraints in your question unless some additional restrictions on $S$ were added.
(please edit or comment for any corrections)