Uno, dos tres,... How many digits are there?

We can use Stirling's Approximation: $\sqrt{2\pi n} \left(\frac{n}{e}\right)^n \le n! \le e\sqrt{n} \left(\frac{n}{e}\right)^n$ Now take the log (base 10) of all these expressions: $\frac{1}{2}\log{(2\pi)} + \frac{1}{2}\log{n} + n\left(\log{n} - \log{e}\right) \le \log{n!} \le \log{e} + \frac{1}{2}\log{n} + n\left(\log{n} - \log{e}\right)$ Now if we substitute $n = 2015$ and the values given, we get the following: $13320.12 \le \log{n!} \le 13320.59$ $13321\le \lfloor \log{n!} \rfloor + 1 \le 13321$ Now the formula for the number of digits in $n!$ is exactly $\lfloor \log{n!} \rfloor + 1$ . Therefore, the number of digits in $2015!$ is $\boxed{13321}$ .