When is $m! > \binom{n}{m}$? 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!
 A: 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:
$$m = - \frac{1}{W{\left( -\frac{1}{e \sqrt{n}} \right) }}$$
