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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?

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  • $\begingroup$ I am not quite sure, but the Steinhauser notation should be an iteration of $n^n$ , exactly what you want. I doubt however that a nice closed form exists. $\endgroup$ – Peter Nov 2 at 13:31
  • $\begingroup$ Do you have a reference for Steinhauser notation? $\endgroup$ – sam wolfe Nov 2 at 14:53
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    $\begingroup$ Maybe this is a good start : sites.google.com/site/largenumbers/home/3-2/Mega $\endgroup$ – Peter Nov 2 at 14:56
  • $\begingroup$ @Peter Could the case $k=n$ be simpler to simplify, somehow? (added as an extra question) $\endgroup$ – sam wolfe Nov 9 at 14:02
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    $\begingroup$ Yes! That's it. $\endgroup$ – sam wolfe Nov 10 at 14:47

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