Babel orders

Relevant wiki: Stirling's Formula

The number we seek is $\large \log(\log((25^{1312000}!))$

We need to make use of Stirling's Approximation and since the number is so huge, any of the approximations will work. I will use $n! \sim \sqrt{2\pi n}(\frac{n}{e})^{n}$ because it doesn't involve logarithms and I plan to use common (base 10) logs.

First rewrite $25^{1312000}$ as $10^{\log{25}\cdot 1312000} = 10^{1834097.291}$ and just call this exponent $A$

The number of orderings is $\sqrt{2 \pi 10^{A}}(\frac{10^{A}}{e})^{10^{A}} = \sqrt{2\pi}\cdot 10^{\frac{A}{2}}(10^{A-\log{e}})^{10^{A}}$

Taking the first common logarithm and separating by the product and power rules results in

$\log{\sqrt{2\pi}}+\frac{A}{2}+10^{A}(A-\log{e})$

It's not possible to take the logarithm of this sum very easily, but note the first term is minuscule and even the second is tiny compared to the third. They can be ignored. Even the $\log{e}$ can be ignored because it is also tiny compared to $A$ . The second logarithm gives the answer.

$A+\log{A} \approx 1834103.555$ which means the number we seek is $1834104$

The number of books can reordered in $25^{1312000}!$ ways.

For very large $n$ , Stirling's approximation is $\ln n! \approx n \ln n - n$ . Using the change of base formula $\ln x = \frac{\log x}{\log e}$ , the equation converts to $\frac{\log n!}{\log e} \approx n \frac{\log n}{\log e} - n$ or $\log n! \approx n(\log n - \log e)$ . This means the approximate number of digits in $n!$ is $n(\log n - \log e)$ , and the number of digits of the number of digits in $n!$ is $\lceil \log (n(\log n - \log e)) \rceil$ = $\lceil \log n + \log (\log n - \log e) \rceil$ .

Therefore, the number of digits of the number of digits in $25^{1312000}!$ is $\lceil \log 25^{1312000} + \log (\log 25^{1312000} - \log e) \rceil$ $=$ $\lceil 1312000 \log 25 + \log (1312000 \log 25 - \log e) \rceil = \boxed{1834104}$ .