Is nᵐ>mⁿ if m>n? I remember playing with my calculator when I was young. I really liked big numbers so I'd punch big numbers like $20^{30}$ to see how big it really is.
On such a quest, I did observe that $20^{30}$ is greater than the value of $30^{20}$. In fact, in many cases, I found that $n^m>m^n$ if $m>n$.
Is this a general fact? If so, can it be proved?

In simple terms, for integers you can start with smallest no. i.e. (1,2), (2,3), (2,4).


*

*$1^2 < 2^1$,

*$2^3 < 3^2$ and

*$2^4 = 4^2$.
In all the above cases $n^m > m^n$ was false for all m>n.
By observing the pattern for all n>=2 and m>4 we have $n^m > m^n$ true. Consider 


*

*(2,5) $\implies$ 32 > 25 or 

*(3,4) $\implies$ 81 > 64 or

*(4,100) $\implies$ (1.6 * 10^60) > 100000000 and so on...
So basically even a small number but with large exponent/power is greater than a big number with small exponent as observed above, except for some cases. Bigger exponent matters more than a big base number.
As for the proof part you can take log of $n^m$ and $m^n$.
As the function $\frac{\log x}{x}$ is a decreasing function for x > e($\approx$ 2.718)  
$\implies$        $\frac{\log n}{n} > \frac{\log m}{m}$    (for m>n)  
So, (as mentioned in above answers also)
m > n > e $\implies$ $n^m > m^n$.
