How to get $e^{\sqrt{\log (x)}} \leq e^{log(x)}=x \leq x^n$?

Question

Hi i was browsing through various asynptotic questions and got stuck in the mid due to the following daubt in the answer given in the link: Prove that $e^{\sqrt{\log x }}=O(x^n)$.

How beni got: $$e^{\sqrt{\log (x)}} \leq e^{\log(x)}=x \leq x^n?$$

Answer 1

That only holds true for large $ x $. For large $ x $, $ \log x > 1 $ which implies that $ \sqrt{\log x} < \log x $ as $ \sqrt{u} < u $ is true for all $ u > 1 $.

The exponential function is increasing, i.e. $ e^u < e^y \iff u < y $ and hence $ e^{\sqrt{\log x}} < e^{\log x} = x $.

Also, for $ x > 1 $, $ x < x^n $ as $ 1 < x \cdot \left(x^{n-1} - 1\right) $. Putting these together yields the result.