Factorial of a large number

To solve this task you need the rules of calculation for the logarithm: $\begin{aligned} \log(\exp(x)) &= x \\ \log(x \cdot y) &= \log(x) +\log(y) \\ \log(x^a) &= a \log(x) \\ \log_a(x) &= \frac{\log(x)}{\log(a)} \end{aligned}$ By logarithmising the Stirling formula it translates into a usable form, that does not cause any overflow errors in the computation: $\begin{aligned} N! &\approx N^N e^{-N} \sqrt{2\pi N} \\ \log(N!) &\approx \log(N^N) + \log( e^{-N}) + \log( \sqrt{N}) +\log( \sqrt{2\pi})\\ &= N \log(N) - N + \frac{1}{2} \log(N) + \log(\sqrt{2\pi}) \\ \log(10^{23}!) &\approx 23 \cdot 10^{23} \log(10) - 10^{23} + 11.5 \log(10) + \log(\sqrt{2\pi}) \\ &\approx 23 \cdot 10^{23} \log(10) - 10^{23} + \mathcal{O}(10) \\ \log_{10}(10^{23}!) = \frac{\log(10^{23}!)}{\log(10)} &\approx 10^{23} \left(23 - \frac{1}{\log(10)}\right) \\ X = \log_{10}(\log_{10}(10^{23}!)) &\approx 23 + \frac{1}{\log(10)}\log\left(23 - \frac{1}{\log(10)}\right)\\ &\approx 24.35 \end{aligned}$

P.S.: I just had to realize, that Wolfram Alpha delivers the finished solution directly, if you only type "(10^23)!" in the command line. That makes me a little sad.