When is $m! > \binom{n}{m}$?

Question

Given $n$ a postive integer, when is $m! > \binom{n}{m}$? I'm looking for an upper bounds on the value of $m$, as a function of $n$, with a formula that hopefully doesn't contain any special functions. Hopefully there is a good upper bounds that meets these criteria!

Answer 1

From Wikipedia's Factorial page->Rate of growth and approximations for large $n$, we have:

$$\tag{1} e \left( \frac{m}{e} \right)^m \le m!$$

...and if we do some simple math:

$$\binom{n}{m} = \frac{n!}{(n-m)!} \frac{1}{m!} \le n^m \frac{1}{m!}$$

Then, plugging in $(1)$ for $m!$:

$$n^m \left( \frac{1}{m!} \right) \le n^m \left( \frac{(e/m)^m}{e} \right) = \frac{(e \cdot n)^m}{e \cdot m^m}$$

...which gives us:

$$\tag{2} \binom{n}{m} \le \frac{(e \cdot n)^m}{e \cdot m^m} \le e \left( \frac{m}{e} \right)^m \le m!$$

So we'd like to find the smallest $m$ such that:

$$\begin{align} \frac{(e \cdot n)^m}{e \cdot m^m} &\le e \left( \frac{m}{e} \right)^m\\ \frac{e^{2m}n^m}{e \cdot m^{2m}} &\le e \\ \frac{e^{2m}n^m}{m^{2m}} &\le e^2 \\ \left( \frac{e \sqrt{n}}{m} \right)^{2m} &\le e^2 \\ \frac{e \sqrt{n}}{m} &\le e^{1/m} \\ e \sqrt{n} &\le m \cdot e^{1/m} \end{align}$$

This is almost the "Lambert $W$ function", or "omega function" of $\sqrt{n}$, which has fairly sizable entries in Wikipedia's page on it and Wolfram's page on it, among other locations.

According to Wolfram Alpha, the corresponding solution, when there is equality, is:

$$m = - \frac{1}{W{\left( -\frac{1}{e \sqrt{n}} \right) }}$$