How is $(\log n)!$ a member of $\theta (\log n \log\log n)$?

Question

Can you please explain how $(\log n)! = \theta((\log n)*(\log \log n))$? I myself used the equation $n! = \theta(n^n)$ and replaced $\log n$ for $n$ as below $(\log n)! = \theta((\log(n))^{\log(n)})$. But it is different. Why shouldn't we use this equation?

Answer 1

In addition to following @Anthony's advice, notice the following: $$ f_n = \Theta(g_n) \Rightarrow \\ (1) \lim_n |\frac{f_n}{g_n}| \geq c >0 \\ (2) \lim_n |\frac{f_n}{g_n}| \leq c < \infty $$ Both of these conditions must fulfill, otherwise you have different orders. Also notice, $$ (\log n)! = \prod_{k=2}^n \log k = e^{\sum_{k=2}^{n} \log \log k} \geq e^{\int_{2}^{n}\log \log x dx} $$ For this integral, after a bit of algebra, you get $$ n \log \log n - \int_{2}^{\log n}\frac{e^t dt}{t} $$ The second integral is an Exponential integral, you should find the solution yourself, but, given the upper bound, it should $\Theta(n)$. Can you handle from here?