# Need help understanding tightest lower bound ( BigOmega ) of n!

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!?

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

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