How many digits do I really have?

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 $n = 10^7$ and the values given, we get the following: $6567059.08 \le \log{n!} \le 6567059.12$ $6567060 \le \lfloor \log{n!} \rfloor + 1 \le 6567060$ Now the formula for the number of digits in $n!$ is exactly $\lfloor \log{n!} \rfloor + 1$ . Therefore, the number of digits in $10^7!$ is $\boxed{6567060}$ .

At first I tried to approximate the given factorial into an integral after taking logarithm (since $10^7$ is very large). But I was off by $4$ , since the approximation wasn't good enough. So I kind of cheated and got the answer using a C++ code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

#include<fstream.h>
#include<conio.h>
#include<math.h>
#include<iomanip.h>

int main()
    {
    double s=0;
    for(double i=2;i<=10000000;i++)
        s+=log(i);
    cout<<setprecision(15)<<floor((s/log(10))+1);
    getch();
    return 0;
    }

Is there any other way to do this other than the solutions already mentioned here?