# Magnitude of n-th factorial

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:

$$3^{5^{5^5}}=10^{10^{10^a}}$$

$$5^{50\times5^{5^5}}=10^{10^{10^b}}$$

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

$$a\approx2.7$$

$$b\approx2.7$$

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:

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

• 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! – Xi'an Mar 8 '17 at 14:28
• Yes, the difference of a few numbers do not really matter when placed in the right places. – Simply Beautiful Art Mar 8 '17 at 14:36
• I am not sure how much it matters but a closer approximation is$$9\underbrace{!...!}_{n\text{ times}}=\underbrace{3^{3^{\vdots^{3^9}}}}_{n\ 3's}$$ – Xi'an Mar 9 '17 at 9:21
• Depends what your doing with the number. If you need closer approximation for tighter bounds, then sure. – Simply Beautiful Art Mar 9 '17 at 13:34