What is (319!)!?

Question

This question came to my mind when someone wrote 319!! on my dorm (room 319). I was able to find 319! and it's about 10^661. I used mathematica to compute (n!)! and I was able to go up to 12 and came out around 10^(3,000,000,000). I would like to know how big (319!)! is, I just don't know where to go from here.

Answer 1

Stirling’s formula says:

$$n!\sim \sqrt{2\pi n}\left(\frac ne\right)^n$$

or $$\log_{10} n!\approx n\log_{10}n-\frac{n}{\ln 10}+\frac{\log_{10}(2\pi n)}{2}$$

When $n$ is huge, the last term is irrelevant.

So $$ \begin{align} \log_{10}\left(10^{661}!\right)&\approx661\cdot 10^{661}-\frac{1 }{\ln 10}\cdot 10^{661}\\&\approx 6.606\cdot 10^{663} \end{align} $$

More generally, $$\log_{10}(10^m!)\approx (m-\log_{10}e)\cdot 10^m$$

Answer 2

If you use the first values you had and plot them, you will notice that $$\log\Big[\log\big[(n!)!\big]\Big]\sim a n - b$$

A quick and dirty regression gives with $R^2=0.999777$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 2.24620 & 0.02429 & \{2.18676,2.30565\} \\ b & 4.99077 & 0.15869 & \{4.60247,5.37906\} \\ \end{array}$$ that we can approximate as $$\log\Big[\log\big[(n!)!\big]\Big]\sim n \log(10) - 5$$ $$\log\big[(n!)!\big]\sim 10^n e^{-5}$$

So $$(319!)! \sim \exp\big[7 \times 10^{216}\big]$$