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    <p>You could estimate upper and lower bounds for <code>(log(n))!</code> using the identity <img src="https://latex.codecogs.com/gif.latex?x%5E%7Blog%28y%29%7D%3Dy%5E%7Blog%28x%29%7D" alt="x^log(y)=y^log(x)"> together with product estimations.</p> <p>For an upper bound:</p> <blockquote> <p><img src="https://latex.codecogs.com/gif.latex?%28log%28n%29%29%21%20%5C%3A%20%5Capprox%20%5C%3A%20%5Cprod_%7Bk%3D1%7D%5E%7Blog%28n%29%7Dk%20%5C%3A%20%5Cle%20%5C%3A%20%5Cprod_%7Bk%3D1%7D%5E%7Blog%28n%29%7Dlog%28n%29%20%5C%3A%20%3D%20%5C%3A%20log%28n%29%5E%7Blog%28n%29%7D%20%5C%3A%20%3D%20%5C%3A%20n%5E%7Blog%28log%28n%29%29%7D" alt="upper bound"></p> </blockquote> <p>For a lower bound:</p> <blockquote> <p><img src="https://latex.codecogs.com/gif.latex?%28log%28n%29%29%21%20%5C%3A%20%5Capprox%20%5C%3A%20%5Cprod_%7Bk%3D1%7D%5E%7Blog%28n%29%7Dk%20%5C%3A%20%5Cge%20%5C%3A%20%5Cprod_%7Bk%3D%5Cfrac%7Blog%28n%29%7D%7B2%7D%5Cright%7D%5E%7Blog%28n%29%7Dk%20%5C%3A%20%5Cge%20%5C%3A%20%5Cprod_%7Bk%3D%5Cfrac%7Blog%28n%29%7D%7B2%7D%5Cright%7D%5E%7Blog%28n%29%7D%7B%5Cfrac%7Blog%28n%29%7D%7B2%7D%7D%20%5C%3A%20%3D%20%5C%3A%20%5Cleft%28%5Cfrac%7Blog%28n%29%7D%7B2%7D%5Cright%29%5E%7B%5Cfrac%7Blog%28n%29%7D%7B2%7D%7D%20%5C%3A%20%3D%20%5C%3A%20%5Cleft%28%5Cfrac%7Blog%28n%29%7D%7B2%7D%5Cright%29%5E%7Blog%5Csqrt%7Bn%7D%7D%20%5C%3A%20%3D%20%5C%3A%20n%5E%7B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20log%5Cleft%28%7B%5Cfrac%7Blog%28n%29%7D%7B2%7D%7D%5Cright%29%7D%20%5C%3A%20%3D%20%5C%3A%20n%5E%7B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20log%28log%28n%29%29-log%5Csqrt%7B2%7D%7D" alt="lowerbound"></p> </blockquote> <p>Combined you will get:</p> <blockquote> <p><img src="https://latex.codecogs.com/gif.latex?n%5E%7B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20log%28log%28n%29%29-log%5Csqrt%7B2%7D%7D%20%5Cle%20%28log%28n%29%29%21%20%5Cle%20n%5E%7Blog%28log%28n%29%29%7D" alt="combined"></p> </blockquote> <p>So at least:</p> <blockquote> <p><img src="https://latex.codecogs.com/gif.latex?%28log%28n%29%29%21%20%5Cin%20O%5Cleft%28n%5E%7Blog%28log%28n%29%29%7D%5Cright%29" alt="O(n^{log(log(n))})"></p> </blockquote> <p>Obviously, the (in)equations are somehow 'odd' due to the non-integer index boundaries of the products.</p> <p><strong>Update:</strong> The bound given by <a href="https://stackoverflow.com/users/535615/hwlau">hwlau</a> using the sterling approximation is lower (by <code>sqrt(log(n))/n</code>) and should be tight.</p> <blockquote> <p><img src="https://latex.codecogs.com/gif.latex?%28log%28n%29%29%21%5Capprox%20%5Cfrac%7B1%7D%7Bn%7Dlog%28n%29%5E%7Blog%28n%29&plus;%5Cfrac%7B1%7D%7B2%7D%7D%20%3D%20%5Cfrac%7B%5Csqrt%7Blog%28n%29%7D%7D%7Bn%7Dlog%28n%29%5E%7Blog%28n%29%7D%20%3D%20%5Cfrac%7B%5Csqrt%7Blog%28n%29%7D%7D%7Bn%7Dn%20%5E%20%7Blog%28log%28n%29%29%7D" alt="tighter bounds"></p> </blockquote>
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    1. POHow can we expand (logn)! and (logn!)! to determine the complexities?
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    1. COJust a reminder, your case is a loose upper bound compared with mine. $$N^(ln(ln N)) = O( lnN^(lnN+1/2)/n )$$
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    2. COYeah that's true. PS: It must be Ω instead of O in your comment.
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    3. COYup, you are right. It is hard to type latex here
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