In simple terms, for integers you can start with smallest no. i.e. (1,2), (2,3), (2,4).
- 1^2 < 2^1$1^2 < 2^1$,
- 2^3 < 3^2$2^3 < 3^2$ and
- 2^4 = 4^2$2^4 = 4^2$.
In all the above cases n^m > m^n$n^m > m^n$ was false for all m>n. By observing the pattern for all n>=2 and m>4 we have n^m > m^n$n^m > m^n$ true. Consider
- (2,5) \implies$\implies$ 32 > 25 or
- (3,4) \implies$\implies$ 81 > 64 or
- (4,100) \implies$\implies$ (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$n^m$ and m^n$m^n$.
As the function \frac{\log x}{x}$\frac{\log x}{x}$ is a decreasing function for x > e(\approx$\approx$ 2.718)
\frac{\log n}{n} > \frac{\log m}{m}
$\implies$ $\frac{\log n}{n} > \frac{\log m}{m}$ (for m>n)
So
So, (as mentioned in above answers also)
nm > mn > e \implies m^n > n^m$\implies$ $n^m > m^n$.