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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.

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    $\begingroup$ You can use Stirling's formula to get an approximation, but computing an exact answer is literally impossible; it is too big. $\endgroup$
    – MJD
    Oct 15 at 17:00
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    $\begingroup$ Notation note: Sometimes $n!!$ is written for $n(n-2)(n-4)\cdots,$ ending at $2$ or $1.$ So $$319!!=1\cdot 3\cdot 5\cdots 317\cdot 319$$ en.wikipedia.org/wiki/Double_factorial?wprov=sfti1 $\endgroup$ Oct 15 at 17:06
  • $\begingroup$ Using the double factorial notation: $$319!!=\frac{319!}{2^{318/2}(318/2)!}$$ $\endgroup$ Oct 15 at 17:08
  • $\begingroup$ But using Sterling’s formula (estimating $n!$) is the best you can do for $(319!)!,$ I’d guess, and even then, it’s not clear how close you can do. $\endgroup$ Oct 15 at 17:10
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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, $$\log_{10}(10^m!)\approx (m-\log_{10}e)\cdot 10^m$$

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  • $\begingroup$ In case Nicholas Hernandez is not familiar with the $\log_{10}$ function: $\log_{10}(n)$ is, roughly, a bound on the number of digits one uses to write the numeral for $n$. So for example $\log_{10}$ of a 1-digit number will be somewhere between 0 and 1, and $\log_{10}(10000000000) = 10$. Thomas Andrews’ result above shows that, if written out, $319!!$ would have something like $6·10^{663}$ digits. For comparison, there are believed to be around $10^{80}$ elementary particles in the universe. $\endgroup$
    – MJD
    Oct 15 at 18:40
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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 $$(319!)! \sim \exp\big[7 \times 10^{216}\big]$$

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