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What ist the time complexitiy of an algorithm with the running time of $a^{n^b}; a,b>1$? And how is it compared to factorial complexity O(n!)?

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  • $\begingroup$ What do you mean by the running time of an expression? You can compute the running time of an algorithm, but not an expression. What are your thoughts? Please edit the question to clarify your question. $\endgroup$ – D.W. Nov 24 '20 at 17:57
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Let's consider $n!<e\left( \frac{n}{2} \right)^n=e^{n\ln n+1-n\ln2}$. On other hand $a^{n^b}=e^{n^b \ln a}$. Now we should compare $(n\ln n+1-n\ln2)$ to $n^b \ln a$, where right member wins, when $a,b>1$: $$\frac{n\ln n+1-n\ln2}{n^b \ln a}\to 0$$

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