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## Goals of the article

This is an article for my father about a size of large numbers question.

How many digits does the following number have?

\begin{equation}\label{eq:start} x = \left((9^9)!\right)! \end{equation}

To start $$9^9 = 387,420,489$$. Next we need Sterling's approximation, a ubiquitous formula of statistical mechanics [1,2,3].

\begin{equation} \log(z!) = z\log(z) - z \end{equation}

or in log base 10

\begin{equation} \log_{10}(z!) \approx z\log_{10}(z) - z\log_{10}(e) \end{equation}

Next for $$z = 387,420,489$$.

\begin{equation} \log_{10}(z!) \approx 3,158,983,316 \end{equation}

So $$(9^9)!$$ is a number with approximately 3.16 billion digits. What happens when we add another factorial?

\begin{align} z &= 10^{3,158,983,316}\\ \log_{10}(z!) &\approx z\log_{10}(z) - z\log_{10}(e)\\ \end{align}

Therefore, if $$b = 3.158983316$$, then

\begin{align} \log_{10}\left[\left((9^9)!\right)!\right] &\approx 10^{b\times 10^9} \log_{10} \left(10^{b\times 10^9} \right) - 10^{b\times 10^9}\log_{10}(e)\\ \log_{10}\left[\left((9^9)!\right)!\right] &\approx 10^{b\times 10^9} \left( \log_{10} \left(10^{b\times 10^9} \right) - \log_{10}(e)\right)\\ \log_{10}\left[\left((9^9)!\right)!\right] &\approx 10^{b\times 10^9} b\times 10^9\\ \log_{10}\left[\left((9^9)!\right)!\right] &\approx b \left(10^{b\times 10^9+9}\right)\\ \log_{10}\left[\left((9^9)!\right)!\right] &\approx 3.158983316 \times 10^{3,158,983,325}\\ \end{align}

So very roughly, $$\left((9^9)!\right)!$$ is a number with $$3\times 10^\text{3 to 3.2 billion}$$ digits.

## References

 D. A. McQuarrie, "Introduction and Review," in Statistical Thermodynamics, pp. 1—34, 2000.

 K. A. Dill, and S. Bromberg, "Entropy and the Boltzmann Law," in Molecular Driving Forces, pp. 81—92, 2011.

 M. S. Shell, "Equilibrium and Entropy," in Thermodynamics and Statistical Mechanics: An Integrated Approach, pp. 6—20, 2015.