Nested Tetration properties Regarding tetration, I know properties like ${}^a({}^bn)= {}^{ab}n$ do not hold in general. When $a=b=2$, for instance, we have
$$
{}^2({}^2n)={}^2\left(n^n \right)=\left(n^n \right)^{\left(n^n \right)}=\left(n^{n^n} \right)^n=({}^3n)^n.
$$
I wonder, however, if there is a way of simplifying the following nested tetration into a closed expression, with $f(n)={}^2n$,
$$
f^k(n)={}^{\overbrace{2\, \cdots\,2 }^{k\text{ times}}}n.
$$
For example, the first five terms are
\begin{align}
f^0(n)&=n\\
f^1(n)&={}^2n=n^n\\
f^2(n)&={}^2({}^2n)=n^{n^nn}\\
f^3(n)&={}^2({}^2({}^2n))=n^{n^{n^nn}n^nn}\\
f^4(n)&={}^2({}^2({}^2({}^2n)))=n^{n^{n^{n^nn}n^nn}n^{n^nn}n^nn}.
\end{align}
I can't find a relatively simple way of simplifying higher order terms. Any ideas?
Extra question: Could the case $k=n$ be "simpler" to simplify, somehow?
 A: No, this does not simplify. It is notable that the specific case of $k=n$ is given by the Steinhaus-Moser notation by definition, but it does not give any other nice forms.
For the general case though, we have the bounds:
$${}^{a+b-1}n\le{}^a({}^bn)\le{}^{a+b}n$$
as an extended version of Knuth's arrow theorem and in your specific case:
$${}^{k+1}n\le f^k(n)\le{}^{k+2}n$$
for sufficiently large $n$.
The lower bound is easy to deduce by noting that:
$$f^{k+1}(n)=f^k(n)^{f^k(n)}\ge n^{({}^{k+1}n)}={}^{k+2}n$$
while the upper bound is deduced by instead proving the even tighter bound:
$$f^k(n)\le\underbrace{n~\widehat~~n~\widehat~~\dots~\widehat~~n~\widehat~~n}_k~\widehat~~(n+k)\le{}^{k+2}n$$
which gives
\begin{align}f^{k+1}(n)&=f^k(n)^{f^k(n)}\\&\le\underbrace{n~\widehat~~n~\widehat~~\dots~\widehat~~n}_k~\widehat~~(n^{n+k}+n+k)\tag{*}\\&\le\underbrace{n~\widehat~~n~\widehat~~\dots~\widehat~~n}_k~\widehat~~(n^{n+k}+n^{n+k})\\&=\underbrace{n~\widehat~~n~\widehat~~\dots~\widehat~~n}_k~\widehat~~(n^{n+k}\cdot2)\\&=\underbrace{n~\widehat~~n~\widehat~~\dots~\widehat~~n}_k~\widehat~~(n^{n+k}\cdot n)\\&=\underbrace{n~\widehat~~n~\widehat~~\dots~\widehat~~n}_k~\widehat~~(n^{n+k+1})\\&=\underbrace{n~\widehat~~n~\widehat~~\dots~\widehat~~n~\widehat~~n}_{k+1}~\widehat~~(n+k+1)\end{align}
where $(*)$ comes from pushing all of the exponents upwards using
$$a^bc\le(a^b)^c=a^{bc}$$
