calculus - Some questions regarding the motivation and use of Stirling's formula

For a while I've been looking for a proof of Stirling's formula and after recently asking about it here and reading about it here, I finally found a proof that wasn't too difficult here. I've got a few questions regarding this formula.

The proof I settled for was not particularly pretty in the sense that it all boiled down to cumbersome verifications of equalities and inequalities. In addition to that it didn't provide an intuitive motivation for why the formula ought to be correct. So without the necessity of rigour, how does one provide a motivation for why the formula ought to be correct?

A while ago I asked a question regarding asymptotic functions and their interchangeability when calculating limits here. I see now that either the answer I got was insufficient or I misunderstood it because I was left with the impression that $n!$ could be replaced by Stirling's formula whenever the limit of some fraction was computed. However, given that $e=\lim_{n\to\infty}n/(n!)^{\frac{1}{n}}$ we have that $n!\sim\left(\frac{n}{e}\right)^n$. So $$\lim_{n\to\infty}\frac{1}{\sqrt{2\pi n}} = \lim_{n\to\infty}\frac{1}{\sqrt{2\pi n}}\frac{n!}{n!} = \lim_{n\to\infty}\frac{1}{\sqrt{2\pi n}}\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{\left(\frac{n}{e}\right)^n} = 1$$ which is evidently incorrect. So under what conditions can I replace $n!$ with Stirling's formula and other functions with such asymptotic properties? This is important to know because otherwise I might end up with nonsense like I did above but in a bit less evident situations. Can I only ever replace all $n!$ with one particular function that is asymptotically equal to it? How does one prove something like that?

Answer

You are correct that Stirling is sometimes used to yield answers without full justification.

Let's review: The expression "$a_n \sim b_n$ as $n\to \infty$" means

$$\lim_{n\to \infty} \frac{a_n}{b_n} = 1.$$

That's it, no more, no less. So Stirling says

$$\lim_{n\to \infty}\frac{n!}{(n/e)^n\sqrt {2\pi n}} = 1.$$

Another way of putting it is

$$\tag 1 n!= (n/e)^n\sqrt {2\pi n}\cdot c_n,\,\text { where } c_n \to 1.$$

$(1)$ is what I always use when I apply Stirling to get at some other limit. For example,

$$\tag 2\frac{n}{(n!)^{1/n}} = \frac{n}{[(n/e)^n\sqrt {2\pi n}\,c_n]^{1/n}} = \frac{e}{(\sqrt {2\pi n}\, c_n)^{1/n}}.$$

One still needs to verify that $(\sqrt {2\pi n}\,c_n)^{1/n} \to 1$ to see $(2)\to e!$ That is the kind of justification you sometimes see left out when someone is using Stirling.