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I am currently learning complexity theory and wasn't able to find a tightest lower bound to BigOmega(n!), I am quite certain it isn't n^n and so wasn't able to reach to a tightest lower bound, can log(n)^n be the one, of course lose lower bounds are possible, but can you please help me understand the tightest lower bound of n!?

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    $\begingroup$ Why you are not trying to use Stirling approximation formula? $\endgroup$
    – OmG
    Jul 6, 2021 at 13:10
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    $\begingroup$ The tight lower bound is $n! = \Omega(n!)$. $\endgroup$ Jul 6, 2021 at 13:29
  • $\begingroup$ @YuvalFilmus I don't feel like that's particularly helpful $\endgroup$
    – exfret
    Jul 9, 2021 at 15:03

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You are indeed correct that it isn't $n^n$ (though this is good enough for most purposes). As OmG mentioned in the comments, there is a standard approximation formula called "Stirling's approximation formula" which gives the following asymptotic bound:

$$n!=\Theta\Big(\sqrt{n}\frac{n^n}{e^n}\Big)$$

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  • $\begingroup$ Big O is an upper bound. OP was looking for a lower bound. $\endgroup$ Jul 9, 2021 at 18:33
  • $\begingroup$ You are right. I will change it to theta to be more precise. $\endgroup$
    – exfret
    Jul 9, 2021 at 23:09
  • $\begingroup$ No I was not aware of Stirling's formula, thanks. $\endgroup$ Jul 12, 2021 at 13:27

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