TITLE: Elementary proof of $m^n\neq n^m$ for almost all natural numbers $m\neq n$
QUESTION [23 upvotes]: $2^4=16=4^2$. In fact, $\{2,4\}$ is the only pair of natural numbers with that property, i.e. if $m<n$ are natural numbers and $m^n=n^m$, then $m=2$ and $n=4$. 
This is easily seen with some analysis: For $m,n\in\mathbf{N}\backslash\{0\}$, the equation $m^n=n^m$ is equivalent to $\sqrt[m]{m}=\sqrt[n]{n}$. By calculus, we can show that the real function  $t\mapsto \sqrt[t]{t}$ is strictly increasing for $t<e$ and strictly decreasing for $t>e$. So the smaller of the two numbers has to be $<e$ and the proposition follows.
My question: Is there an elementary proof? By elementary I mean most of all no irrational numbers, no calculus.

REPLY [4 votes]: There is a rather pretty picture: given any real numbers $a,b > 1,$ the relation $$ a^b = b^a  $$ is equivalent to
$$ \frac{\log a}{a} =  \frac{\log b}{b}.$$
So what you do is draw the graph 
$$ y = \log x  $$ and then draw any line through the origin with positive slope (but smaller than $1/e$). The line intersects the curve in two points, with $x$-values $a,b$ satisfying $ a^b = b^a.$ The smaller of the two numbers $a,b$ lies between $1$ and $e,$ so the only possible integer is $2.$ In that case, the other $x$-value is $4.$