Fun Num

$L = \dfrac{18^{20}}{20^{18}} = \dfrac{2^{20} \times 9^{20}}{2^{18} \times 10^{18}} = 4 \times 10^{2} \times \left(\dfrac{9}{10}\right)^{20} = 400 \times \left( \left(1 - \dfrac{1}{10}\right)^{10}\right)^{2}$ .

Now $f(x) = \left(1 - \dfrac{1}{x}\right)^{x}$ is an increasing function (for $x \gt 1$ ) with limit $\dfrac{1}{e}$ as $x \to \infty$ , so $f(2) = \dfrac{1}{4} \lt \left(1 - \dfrac{1}{10}\right)^{10} \lt \dfrac{1}{e}$ , and thus

$1 \lt \dfrac{400}{16} \lt L \lt \dfrac{400}{e^{2}} \Longrightarrow \boxed{18^{20} \gt 20^{18}}$ .