Elementary proof of $m^n\neq n^m$ for almost all natural numbers $m\neq n$ $2^4=16=4^2$. 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.
 A: HINT $\ $ wlog $\rm\displaystyle\ m > n\ \Rightarrow\ \bigg(\frac{m}n\bigg)^n =\: n^{m-n}\in \mathbb Z\ \Rightarrow\ k := \frac{m}n \in\mathbb Z\ \Rightarrow\ k = n^{k-1}\ \Rightarrow\:\cdots$
NOTE $\ $ Above I have implicitly invoked the Rational Root Test to infer that for $\rm\:j\in \mathbb Z,\: n\in \mathbb N\:,\:$ rational roots of $\rm\ x^n -j\ $ must be integers. $\:$ Above is the special case $\rm\ x = m/n,\ j = n^{m-n}\:.$
A: Have a look here (in particular from "Try this:  Substitute").
A: 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 $4.$
A: The relation $m^n=n^m$ implies that $n,m$ have the same prime factors $p_1<p_2<...<p_k$. Suppose $m=\prod p_i^{\alpha_i},\ n=\prod p_i^{\beta_i}$. By the unique factorization theorem and the relation $m^n=n^m$ we get that $\alpha_i n=\beta_i m$. Suppose $m>n$. This implies $\alpha_i>\beta_i$ since $\alpha_i/\beta_i=m/n$. Therefore $n|m$.
Denote $m=dn$ and $(dn)^n=n^{(dn)}$ i.e. $dn=n^d$ or $d=n^{d-1}$. For $n \geq 2$ we have $n^{d-1}\geq d$ with equality for $d=1$ (which is not good since $m>n$) or $d=2,n=2$ and therefore $m=4$.
