Time complexity of $a^{n^b}; a,b>1$
Let's consider $n!&lt;e\left( \frac{n}{2} \right)^n=e^{n\ln n+1-n\ln2}$. 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 $a,b&gt;1$:
$$\frac{n\ln n+1-n\ln2}{n^b \ln a}\to 0$$