TITLE: Comparing $2^n$ to $n!$
QUESTION [12 upvotes]: Comparing $\;2^n\;$ and $\;n!\;$
I need to rewrite one or the other so they can be more easily compared (i.e. have similar form). Because of the factorial, I'm a little lost as to how to compare the two functions. Normally, I would take the logarithm when trying to re-express a power like $2^n$, but I don't think this will work, since I don't know how to take the logarithm of a factorial, if at all. 
Might the series expansions of either functions be the right approach? If not could, someone point me in the right direction?

REPLY [3 votes]: Method 1: You can take a graphical approach to this problem:

It can be seen that the graphs meet at (0, 1), $2^x$ is greater until they intersect when $x \approx 3.459$, and then the factorial becomes much greater. Alternatively, plot $x! - 2^x$ to see a demonstration of the difference.
Method 2: use Stirling's approximation,
$$x! \sim \left(\frac{x}{e}\right)^x\sqrt{2\pi{x}}\left(1 + \frac{1}{12x}\right)$$
Now you can take logarithms of both expressions. Comparing the graphs of $y = x$ and the base 2 logarithm of the above expression, it can be seen that the latter quickly rises above the former.