I would like a proof that if $a<b$ are integers with $2\leq a,b$ we have $a^b>b^a$ unless $a=2,b=3$ or $a=2,b=4$ . I would like to use as little calculus as possible. Here is my current solution:
Case 1: $a>2$
Fix $a$ and start with $b=a$. notice $a^a=a^a$. Now increase $b$ by $1$ to now get $a^{a+1}>(a+1)^a$. When $a$ is at least three this is true because the left side was multiplied by $a$ and the right side by $(1+\frac{1}{a})^a$.
Since $(1+\frac{1}{a})^a=\sum\limits_{i=0}^a \binom{a}{i} \frac{1}{a^i}$ is the sum of $a+1$ elements all of them smaller than $1$ and the last two terms add $\binom{a}{a-1}\frac{1}{a^{a-1}}+\binom{a}{a}\frac{1}{a^a}=\frac{1}{a^{a-2}}+\frac{1}{a^a}<\frac{1}{3}+\frac{1}{27}<1$ we conclude the sum is less than $a$. Therefore the inequality holds. Notice that each time we add $1$ to $b$ the left side is multiplied by $a$ and the right by something smaller than $(1+\frac{1}{a})^a$, hence the inequality holds for all $b$ larger than $a$.
Case $a=2$ is the same thing only we check $b=3,4$ by hand and then do something similar.
I am not really happy with the current solution, can we find something simpler? If it can be more combinatorial it would be better. Notice I want as little calculus or inequalities as possible.