In connection with a riddle on The Riddler, I would like to know how to evaluate even crudely the order of magnitude of an iterated factorial like $$(\ldots(9\underbrace{!)!\ldots)!}_{n\text{ factorials}}$$ Using Stirling's approximation does not get me very far: \begin{align*} 9! &\approx 3^9\\ (9!)! &\approx (3^9/3)^{3^9}=3^{8\times 3^9}\\ &... \end{align*}


First, let me explain why the numbers in a power tower don't matter. Let us compare these two:



Clearly, we will have $a<b$, but by how much? Well, take the log of each thrice, and you will find that



Indeed, the only things that truly matters is how tall the power towers are, which is why may use a crude Stirling approximation:

$$k!\approx k^k$$

Also, a quick symbolization:

$$k!_n=k\underbrace{!!!!\dots~ !}_n$$

And furthermore,

$$k!_n\approx k^{k^{k^{\dots}}}\bigg\}(n+1)\text{ powers}$$

In terms of Knuth's up-arrow notation:

$$k!_n\approx k\uparrow\uparrow(n+1)$$

  • $\begingroup$ Indeed, I had started to realise the additional factors did not matter so much and hence the very first approximation I thought of $9^{9^{9^\vdots}}$ was given a first order solution! $\endgroup$ – Xi'an Mar 8 '17 at 14:28
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    $\begingroup$ Yes, the difference of a few numbers do not really matter when placed in the right places. $\endgroup$ – Simply Beautiful Art Mar 8 '17 at 14:36
  • $\begingroup$ I am not sure how much it matters but a closer approximation is$$9\underbrace{!...!}_{n\text{ times}}=\underbrace{3^{3^{\vdots^{3^9}}}}_{n\ 3's}$$ $\endgroup$ – Xi'an Mar 9 '17 at 9:21
  • $\begingroup$ Depends what your doing with the number. If you need closer approximation for tighter bounds, then sure. $\endgroup$ – Simply Beautiful Art Mar 9 '17 at 13:34

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