Is there a "simple" way of proving Stirlings formula?

Question

Is there any way to derive Stirlings formula that only requires some undergraduate knowledge of calculus, real analysis and perhaps some identitets involving the gamma function, maybe Wallis product, and things along those lines? If not, and I know this is a rather vague question, what is the simplest but still sufficiently rigorous way of deriving it? I had a look at Stirling's formula: proof? but the comments seems quite messy. Wikipedia was not particularly helpful either since I have not learned about Laplace's method, Bernoulli numbers or the trapezoidal rule.

Answer 1

You may find a "simple" or better well written proof in the book Real Analysis and Foundations from Steven G. Krantz (p. 277 and 278) which you may find. As far as I can see, he uses the Gamma function, l'Hopital's rule and the evaluation in polar coordinates of $$\int_{-\infty}^\infty e^{-s^2} ds$$

Answer 2

As an extremely simple derivation of bounds Stirling's approximation sets, notice that

$$\ln(n!)=\sum_{k=1}^n\ln(n)>\int_1^n\ln(x)\ dx=n\ln n-n+1$$

This inequality is simply a right Riemann sum of a monotonically increasing function. Similarly,

$$\frac12\ln(n)+\ln((n-1)!)=\frac12\left(\ln(n!)+\ln((n-1)!)\right)\\=\frac12\sum_{k=1}^n\ln(n)+\frac12\sum_{k=1}^{n-1}\ln(n)<\int_1^n\ln(x)\ dx=n\ln n-n+1$$

Which is the trapezoidal sum (average of left/right Riemann sums). From each of these, you easily get that

$$e\left(\frac ne\right)^n<n!<\sqrt{n+1}\left(\frac{n+1}e\right)^n$$

The lower bound may then be improved using

$$n!=n(n-1)!<e\sqrt n\left(\frac ne\right)^n$$

---

Stronger results that follow this line of thought may be derived by the much better Euler-Maclaurin formula.