Googology Is Always Fun

Using Stirling's Approximation, let's roughly approximate $n!$ as $\left( \dfrac{n}{e} \right)^n \approx n^n$ . Then a factorial of a factorial of $n$ would be

$(n^n)^{(n^n)}$

Letting $n=10^{63}$ , let's look at $n^n$ , which becomes

$(10^{63})^{(10^{63})} = 10^{{63}\cdot 10^{63}} \approx 10^{10^{65}}$

Now we do the factorial twice

$(n^n)^{(n^n)}=(10^{10^{65}})^{10^{10^{65}}}= 10^{10^{65}10^{10^{65}}} = 10^{10^{65+10^{65}}} \approx 10^{10^{10^{65}}}$

Note: Since $e \approx 10^{0.43}$ , and if $n=10^p$ , then $\left( \dfrac{10^p}{10^{0.43}} \right)^{10^p} = \left( 10^{p-0.43} \right)^{10^p} \approx (10^p)^{(10^p)}$ .

Sterling's approximation states that $n!\approx \sqrt{2\pi n}(\frac{n}{e})^n$ . But that's tedious (considering the large numbers you have here, since no one wants to raise numbers to the power of a vigintillion and take their square root and all that), so we can simplify that. On this scale, $n!$ is slightly larger than $10^{n}$ ; thus, $((10^{63})!)!$ is slightly larger than $10^{10^{10^{63}}}$ , so $10^{10^{10^{65}}}$ would be the best choice here.

What topic would this be? There is no "googology section", so what could it be? Geometry? Definitely not. Algebra? Maybe. Computer science? Probably not.