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Being familiar with Stirling's formula for factorial: $$n!\sim\sqrt{2\pi n}\left(\frac n e\right)^n,\quad\color{gray}{n\to\infty}$$ I naïvely assumed that for twice iterated factorial we can simply substitute the right-hand side into itself and write $$(n!)!\,\stackrel{\color{red}{\small\text{wrong}}}\sim\,\sqrt{2\pi \sqrt{2\pi n}\left(\frac n e\right)^n}\left(\frac{\sqrt{2\pi n}\left(\frac n e\right)^n}e\right)^{\sqrt{2\pi n}\left(\frac n e\right)^n},$$ but, as it turns out, I was wrong. Actually, it grows faster than that: $$(n!)!\,\succ\,\sqrt{2\pi \sqrt{2\pi n}\left(\frac n e\right)^n}\left(\frac{\sqrt{2\pi n}\left(\frac n e\right)^n}e\right)^{\sqrt{2\pi n}\left(\frac n e\right)^n}.$$

Can we express the correct asymptotic growth rate of twice iterated factorial $(n!)!$ using only elementary functions and, possibly, also their inverses, such as the Lambert W-function?

Update: Apparently, the same issue arises for simpler functions like $2^{n!}$. If we simply substitute Stirling's formula for the exponent $n!$, we will get an incorrect asymptotic.

Vladimir Reshetnikov
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    You can use a bound on each side, but they won't agree with one another in the limit. The problem is that if the exponent behaves as $n^k$ then you need to resolve the relative error of the inside of the parentheses down to $n^{-k}$ in order to have bounded relative error overall...and in this case the exponent grows faster than any polynomial. So no finite number of Stirling series terms will get you a correct asymptotic expansion. – Ian Apr 02 '22 at 03:10
  • This answer about inverse factorial might be related: https://math.stackexchange.com/a/461207/19661 – Vladimir Reshetnikov Apr 02 '22 at 07:34
  • https://math.stackexchange.com/a/1587792/815585 – FShrike Apr 02 '22 at 09:22
  • @Ian Is your argument a heuristic, or it can be made rigorous? Does it apply only to a truncated Stirling series, or any combinations of elementary functions? – Vladimir Reshetnikov Apr 02 '22 at 23:04
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    What I said is only for trying to get an asymptotic by taking a Stirling series expression and plug it into itself. No idea about other possibilities. – Ian Apr 03 '22 at 02:39

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