This is just a "share your knowledge" question for a funny infinite-fraction-ish bound I ran into for the inverse factorial function. And to be clear, by "inverse" I don't mean $\frac{1}{n!}$ but rather a funcion $L(n)$ such that $L(n)!\approx n$.
To keep things simple and avoid non-integer gamma stuff, let's define $L(n):\mathbb{N}\rightarrow\mathbb{N}$ to be the unique value such that
$$L(n)!\le n <(L(n)+1)!$$
In other words, $L(n)!$ is the largest factorial not greater than $n$.
What kinds of bounds can we get?
$$f(n)<L(n)\le g(n)$$
Best Answer
If you look here and at related questions and answers, @robjohn gave the exact solution$$\color{blue}{n=\left\lceil e\exp\left(\operatorname{W}\left(\frac1{e}\log\left(\frac{\Gamma(n)}{\sqrt{2\pi}}\right)\right)\right)+\frac12 \right\rceil}$$ where $W(.)$ is Lambert function