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Light problem. I think I have two possible answers: $9^9!$ or is $9^{9!}$ greater ? Unfortunately I haven't been able to find a calculator that can display either answer.

Is there something that is totally different that is larger ?

Happy holidays!

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    How about $9!!$? –  Dec 26 '20 at 17:28
  • The above sounds like I'm very excited. –  Dec 26 '20 at 17:28
  • You can just define $A$ to be an incredibly large number. So no such number exists. – Frederik Ravn Klausen Dec 26 '20 at 17:29
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    @JetChung Did you mean $(9!)!$ ? This is smaller than $9^9!$ because $9!\lt 9^9$. I've also seen double factorial to mean "skip every other number" I.e. $9!!=9\cdot 7\cdot 5\cdot 3\cdot 1$, which is even smaller. –  Dec 26 '20 at 17:30
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    What about $9^{9^9}$? – Infinity_hunter Dec 26 '20 at 17:31
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    Mathematica gives $9^9!\sim 7\times 10^{3158983320}$, making it the largest number mentioned here, by far. – imas145 Dec 26 '20 at 17:32
  • @StinkingBishop ah, you're right. I blame my painkillers... –  Dec 26 '20 at 17:33
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    @Infinity_hunter $9^{9^9}$ is smaller than $9^9!$ because it's the product of $9^9$ nines, while $9^9!$ is the product of $9^9$ numbers, most of which are larger than $9$. –  Dec 26 '20 at 17:34
  • using the gamma function definition of ! we have that -1! $= \infty$ –  Dec 26 '20 at 17:40
  • You have to admit using Gamma is creative. With minimal introduction: 1/0 Or (the triple exponent, that I seem unable to write): ∞^∞^∞ Although mabe infinity is only available on foreign keyboards. I think I remember seeing it on a Selectric Ball on an electric typewriter back in the 70's. – Jim Clark Dec 26 '20 at 18:00
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    Hold down the key and hit the sequence 236 on the numeric keypad and you get ∞ :) – user5713492 Dec 26 '20 at 18:13

4 Answers4

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Using tetration, you can get quite a large number. ${^{n}a}$ is defined as ${{a^a}^a}^{...}$. In other words, it is repeated exponentiation.

Therefore, using just ^, 9, 9 you can get:

${^{9}9}$

which is veeeery big.

Alejandro Breen
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If we allow tetration notation, then $\displaystyle {}^{{}^{9}9}9$ is larger than the other suggestions. For clarity's sake. this means an iterated exponentiation of the form $$9^{9^{9^{.^{.^{.}}}}}$$ where the height of the tower is $$9^{9^{9^{9^{9^{9^{9^{9^{9^{9}}}}}}}}}$$ I'm not sure what this looks like in standard form, however.

David Quinn
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You will need logarithms for this; Stirling's approximation gives $$ \log n! = n \log n -n + \frac{1}{2} \log 2 \pi n + \frac{1}{12n} + \frac{1}{288n^2} - O(n^{-3} )$$ so we get $$ \log 9^{9^9} = 9^9 \log 9,$$

$$ \log 9^9! \approx 9^9 (9 \log 9 - 1 ) $$ which is larger

Put briefly $$ 9^9 \log 9^9 = 9^9 \cdot 9 \log 9 = 9^{10} \log 9 $$

Will Jagy
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Well, $9^9!$ is a product of $9^9$ terms ($1, 2, 3, ...$), $9^9 - 9$ of which are greater than $9$. Since $9^9 - 9 > 9!$, $$9^9! > 9^{9!}$$

Ansar
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