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I know that $n!$ growth faster than $a^n$, (Do factorials really grow faster than exponential functions?).

But what about this: $n!$ and $2^{2^n}$ ($2$ to the power of $2$ to the power of $n$)? Can you please prove?

OmG
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rakamakafo
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  • $2^{2^n}$ grows faster than $n!$ but I reply "What about $2^{2^n}$ vs $(n!)!$?" maybe $2^{2^{2^n}}$. For $n=4$ we have already $19,729$ digits... – Raffaele Sep 19 '17 at 15:07

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Hint: Suppose $n$ times of $2^n$: $$2^n\times\cdots\times2^n = 2^{n^2}$$

It's straightforward to show that $n < 2^n$ and growth of $n$ is less than $2^n$. Therefore, growth of $n!$ is less than $2^{n^2}$. Also $n^2 < 2^n$ and for their growth respectively. Hence, growth of $n!$ is less than $2^{2^n}$.

OmG
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