<s>[INST] Which is larger :: $y!$ or $x^y$, for numbers $x,y$. This is a generalization of this question :: Which is larger? $20!$ or $2^{40}$?.
No explicit general solution was presented there and I'm just curious :D
Thank-you.

Edit :: I want a most-general solution lfor arbitrary $x$ and $y$; not some specfic cases which can then be solved by direct computation. Below, Ahaan-Rungta shows that the case $x < y$ is the one to be considered.
 [/INST] $$\frac{20!}{2^{40}}= \frac{20!}{4^{20}} =
                 \left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right)
\left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)\cdots \left(\frac{20}{4}\right)>\cdots$$
$$ \left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right)
\left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)=\frac{9!}{4^9}=\frac{2835}{2048}>1,$$
so $20!>2^{40}$.
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