Proof that $a^b>b^a$ if $aI 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.
 A: Consider $\mathbb{N}^a$ as a Cartesian lattice. Consider the collection of paths of length $b$ that


*

*start at the origin

*grow in legs of length $1$, moving parallel to one of the $a$ axes

*only grow outward from the origin


There are $a^b$ such paths, since for each leg you have $a$ directional options. We can identify a subset of these paths whose cardinality is greater than $b^a$, proving the claim.
Let $S$ be the subset of such paths where there is one direction which precisely  $b-a$ of the legs run parallel to, another direction with precisely two legs running parallel, and each of the remaining directions has precisely one parallel leg. 
When $b>a+2$, combinatorics tells us that there are $$|S|=\binom{b!}{b-a;2;\overbrace{1;\cdots;1}^{a-2}}\cdot a(a-1)$$ such paths. The multinomial coefficient counts how many paths have $b-a$ legs specifically parallel to the first direction, and two legs specifically parallel to the second. The $a(a-1)$ factor then accounts for all the permutations. Since $b>a+2$, there is no over counting. (We can handle $b=a+1$ and $b=a+2$ similarly, later below).
So $$\begin{align}
a^b
&>\binom{b!}{b-a;2;\overbrace{1;\cdots;1}^{a-2}}\cdot a(a-1)\\
&=b(b-1)\cdots(b-a+1)\cdot\binom{a}{2}\\
\end{align}$$
$b(b-1)\cdots(b-a+1)$ is shy of $b^a$, but the $\binom{a}{2}$ factor should compensate if $a$ is at least $3$. The limit of the ratio $\frac{b(b-1)\cdots(b-a+1)}{b^a}$ is $1$, so this certainly happens for large enough $b$. (It would be nice if we could establish it holds for all $b>a+2$.) Alternatively we could include more paths in $S$ that contribute like $\binom{b!}{b-a-1;2;2;\overbrace{1;\cdots;1}^{a-3}}$ which would cancel the subleading term in $b(b-1)\cdots(b-a+1)$.
When $b=a+2$, $|S|=\binom{b!}{2;2;\overbrace{1;\cdots;1}^{a-2}}\cdot \binom{a}{2}$, and when $b=a+1$, $|S|=\binom{b!}{1;2;\overbrace{1;\cdots;1}^{a-2}}\cdot \binom{a}{1}$. The above parts needs slight modification to compensate.
A: Assume that $n\geq3$. Then
$$\left(1+{1\over n}\right)^n=\sum_{k=0}^n{n\choose k}n^{-k}<\sum_{k=0}^n{1\over k!}<e<n\ .$$
It follows that $(n+1)^n<n^{n+1}$, which implies that the function $$n\mapsto n^{1/n}\qquad(n\geq3)$$
is strictly decreasing. This takes care of all cases with $3\leq a<b$. Furthermore one has
$$2^{1/2}=4^{1/4}>n^{1/n}\qquad(n\geq5)\ ,$$
and this settles the remaining cases.
