How does $a^{a^n}$ compare with $n!$ asymptotically?

How do they compare?: $n!=\mathcal{O}(a^{a^n})$

Proof: If we bring down both $n!$ and $a^{a^n}$ to $log$ scale then $n!$ in $log$ scale will be approximately $nlogn$ (as $log(n!)≈nlogn$ using Stirling's approximation) and $a^{a^n}$ will be brought down to $a^n*loga$.

$nlogn$ grows slower than $a^n*loga$ which implies $n!=\mathcal{O}(a^{a^n})$.

Applying $\ln\circ\ln$ we obtain \begin{align*} \ln\left(\ln\left(a^{a^n}\right)\right) &=\ln\left(a^n\ln a\right)\\ &=\ln a^n+\ln\left(\ln\left(a\right)\right)\\ &=n\ln a+\ln\left(\ln\left(a\right)\right)\\ &\sim n\ln a\tag{1} \end{align*}

Recalling Stirlings formula $n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$ we obtain

\begin{align*} \ln\left(\ln\left(n!\right)\right)&\sim\ln\left(\ln\left(\left(\frac{n}{e}\right)^n\sqrt{2\pi n}\right)\right)\\ &=\ln\left(\ln\left(\frac{n}{e}\right)^n+\ln\sqrt{2\pi n}\right)\\ &=\ln\left(n\ln n-n+\ln\sqrt{2\pi}+\ln n\right)\\ &\sim \ln\left(n\ln n\right)\tag{2} \end{align*}

We conclude from (1) and (2)

\begin{align*} \ln\left(n\ln n\right)=\mathcal{O}\left(n\ln a\right) \end{align*} from which \begin{align*} \ln\left(\ln\left(n!\right)\right)=\mathcal{O}\left(\ln\ln a^{a^n}\right) \end{align*} and finally \begin{align*} \color{blue}{n!=\mathcal{O}\left(a^{a^n}\right)} \end{align*} follows.