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.
 A: For a fixed $x,$ $y!$ will eventually be larger, as you can tell by using Stirling's formula.
A: Hint:
In general, for positive integer $x,y$ we have:
$y!=y(y-1)(y-2) \cdot\cdot\cdot2\cdot1$
$x^y=xxx\cdot\cdot\cdot xx$


*

*If $x\geq y$; then $ x^y\ge y! $

*If $x<y $, ?! (Now this is the question!)

Extra:
If $ \left\lceil{\frac{y}{2}}\right\rceil+1<  x<y$; then $x^y\ge y!.$ (easy proof)
Now, we have only the case $ x\le\lceil{\frac{y}{2}}\rceil +1  $
A: "Just compute them and compare" is the only fully failsafe method.
In most cases, however, estimating the logarithms of both of $x^y$ and $y!$ using Stirling's formula will yield a conclusive result without needing to compute the two values in high detail.
A: $$\frac{20!}{2^{40}}= \frac{20!}{4^{20}} =
                 \left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right)
\left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)\cdots \left(\frac{20}{4}\right)>\cdots$$
$$ \left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right)
\left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)=\frac{9!}{4^9}=\frac{2835}{2048}>1,$$
so $20!>2^{40}$.
