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?
 A: One more attempt:
Consider the function:
$f(x) := \dfrac{x}{\log x}$ , $x \gt e$ (say), is strictly increasing,  since
$f'(x) = \dfrac{\log x - 1}{(\log x)^2} \gt 0$ for $x \gt e$.
$f(x_1) \lt f(x_2)$ for $ x_1 \lt  x_2$.
With  $x_1 = n$ ,  $ x_2= m $,  $  n \lt m $ , $m,n$ positive integers
$\dfrac{n}{\log n} \lt \dfrac{m}{\log(m)}$;
$n \log (m) \lt (m) \log n$;
$\log (m)^n \lt \log (n)^m$ ;
$\exp(\log (m)^n) \lt \exp (\log (n)^m)$;
$m^n \lt n^m$ for $m\gt n.$
A: For a positive integer $m$, consider the function $f(x)=m^x/x^m$. And $g(x)=\ln f(x)=x\ln m-m\ln x$.
Then
$$g'(x)=\ln m-\frac mx$$
which is positive for $x>m/\ln m$. Then $g$ is increasing in $(m/\ln m,\infty)$. For $m>e$ we have $m>m/\ln m$ and $g(x)>0$ for $x>m$. Then, as it has been said in comments,
$$n>m>e\implies m^n>n^m$$
A: You are proposing that $n^m > m^n \iff n > m$. However, there are many examples where this is not entirely true.
If $n = 2 \land m = 3 \implies n^m < m^n : n < m$
If $n = 2 \land m = 4 \implies n^m = m^n : n < m$
And obviously if $n = 1 \land m > 1 \implies n^m < m^n : n < m$
But perhaps what you are trying to say is that:
If $n > m \implies n^m > m^n$ because it seems like $n < m$ in these contradicting examples above. I mean, why do these seem to be the only contradicting examples? With the examples above, we know that $n \neq m \neq 0 \lor 1 \because n^m < m^n$. So moving from $1$ to $2$ where $n = 2$ and $m > 2$, we find a little shift in the equality signs.
For the first example, $n^m < m^n$
For the second example, $n^m = m^n$
And it seems that if $m > 2^2 = 4$ then your theory is true where $n^m > m^n$. And it seems like the reason your theory looks true on the condition that $m > 2^2$ is because we must find the first $n^m \lor m^n : n \land m > 1$ (because $1 > 0$ which is obviously $2^2$).
In summary, the theory is not that "if $n^m > m^n$ then $n > m$" but is instead that:
$$\text{if} \qquad n > m \implies n^m > m^n : n \land m \in \mathbb{W}$$
(since $\mathbb{W}$ is the set of all numbers $\ge 0$ aka "Whole Numbers")
A: 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 $\implies$ $n^m > m^n$.
A: This is not an answer  I just need to show the graph. I forgot to label axis. Horizontal is $n$ and the other is $m$
Hope you like it

