Simplifying a 'fractal-like' expression with tetration Let $f_2(n)=2^n n$ and let $f_3$ be defined recursively as
$$
f_3(n)=\underbrace{f_2\cdots f_2}_{n\text{ times}}(n)=f_2^n(n).
$$
This will lead to tetration, but is it possible to write $f_3$ in a closed formula, using the notation ${^{n}a}$ for tetration (or even Knuth's up-arrow notation)?
I tried to write a simple code to see an emerging pattern, but simplifications seem somehow tricky. Also, ${^{n}2}$ seems to be relatively simple to recognise in the expansion of $f_3$, but the remaining terms are somewhat complicated and it almost feels like the expression behaves like a fractal. Any ideas?
Here are the first four iterations, might help to picture the pattern.
\begin{align*}
f_2(n)&=2^n n\\
f_2^2(n)&=2^{(2^n+1)n}n=2^{2^nn}2^nn\\
f_2^3(n)&=2^{(2^{(2^{n}+1)n}+2^n+1)n}n=2^{2^{2^nn}2^nn}2^{2^nn}2^nn\\
f_2^4(n)&=2^{(2^{(2^{(2^n+1)n}+2^n+1)n}+2^{(2^n+1)n}+2^n+1)n}n
\end{align*}
Just to give some context, these functions were defined and discussed in a recent Numberphile video about $\text{TREE}(n)$ and Graham's number (see around minute $10$).
 A: The simple answer is no, because there's a mix of operations being iterated: either multiplication and exponentiation, or if you bring the multiplication up a level, addition and exponentiation.
As far as approximations go though, you can see here for some tight bounds:
$$n\operatorname{Tet}(2^n,1,n)\le f_3(n)\le\operatorname{Tet}(2\sqrt[n]n,n,n)$$
where $\operatorname{Tet}(a,b,c)=\underbrace{a~\widehat~~a~\widehat~~\dots~\widehat~~a~\widehat~~}_cb$ is $c$ powers of $a$ with $b$ on top.
A: I believe I've found a potential solution. $f_3$ may be expressed as
$$
f_3(n)=\log_2\left(g^{n}\left(2^n \right) \right),
$$
where $g(n)={}^2n$. Indeed, iterating over $g$, which corresponds to the iterations of $f_2$, we have
\begin{align}
f_2(n)&=\log_2\left(g\left(2^n \right) \right)=\log_2\left(2^{2^nn} \right)=2^nn\\
f_2^2(n)&=\log_2\left(g^2\left(2^n \right)\right)=\log_2\left({}^2(2^{2^nn} )\right)=\log_2\left(2^{2^{2^nn}2^nn} \right)=2^{2^nn}2^nn\\
f_2^3(n)&=\log_2\left(g^3\left(2^n \right)\right)=\log_2\left({}^2({}^2(2^{2^nn} ))\right)=\log_2\left(2^{2^{2^{2^nn}2^nn}2^{2^nn}2^nn} \right)=2^{2^{2^nn}2^nn}2^{2^nn}2^nn\\
&\vdots
\end{align}
It is then clear that the only simplification remaining is of the following
$$
g^n(a)={}^{\overbrace{2\,\cdots\, 2}^{n\text{ times}}}a.
$$
While in general it is not true that ${}^a({}^bn)={}^{ab}n$, a closed formula for $g^n(a)$ might be possible. Any comments are appreciated. A follow-up question can be found here.
Note: Even though this is not a full solution, I decided to post it as an answer in order to present my ideas more clearly.
Edit note: Based on a comment by @r.e.s., I simplified a bit my original expression for $f_3$ by noticing that $g(2^n)=2^{2^nn}$.
