How to prove if $e<a<b$ then $a^b>b^a$
Thus far I got:
$a^b>b^a$
$e^{\ln(a^b)}>e^{\ln(b^a)}$
$\frac{e^{b\cdot \ln(a)}}{e^{a\cdot \ln(b)}}>1$
$e^{b\cdot \ln(a)-a\cdot \ln(b)}>1$
$b\cdot \ln(a)>a\cdot \ln(b)$
We also know $b>\ln(b)>\ln(a)>1$ and $b>a>\ln(a)>1$
You tagged your question "precalculus" so here is an answer which doesn't use calculus. But we will use the following fact, which you can prove in various ways, depending on how you define exponentials: for all $x$, $$e^x \geqslant 1+x$$ $$e^x-1 \geqslant x$$ We are given that $a>e$: $$ \log a > 1$$ Multiply the two inequalities (valid provided $x>0$): $$(e^x-1 ) \log a>x$$ Now put $x=\log \frac{b}{a}$ (which must be $>0$ so we need $b>a$): $$\left(\frac{b}{a}-1 \right) \log a>\log \frac{b}{a}$$ $$\frac{b}{a} \log a>\log b $$ $$b \log a>a \log b $$ as required.
$a^b>b^a \Leftrightarrow b \ln a>a \ln b \Leftrightarrow \frac{\ln a}{a}>\frac{\ln b}{b}$ when $e<a<b$
So, it is enough to show that $f(x)= \frac{\ln x}{x} $ is monotonically decreasing when $x>e$. Take the derivative and you can get $f'(x)=\frac{1-\ln x}{x^2}<0$.
Take function $f(x)=x^{\frac 1x}$
$f'(x)=\dfrac{x^{\frac 1x}}{x^2}(1-\ln x)<0$ for $x>e$. So, $f(x)$ is monotonically decreasing for $x\gt e$.
Hence, $e\lt a\lt b \implies a^{\frac 1a}\gt b^{\frac 1b} \implies a^b\gt b^a$.
