In reply to
http://math.eretrandre.org/tetrationforu...73#pid4073
First it is easy to see that for \( 1<b<\eta \):
\( {^n b}=\exp_b^{\circ n}(1)\to a<e \) (\( a \) is the lower fixed point of \( b^x \))
Hence for \( n_0 > 3 \) we have for all \( n\ge n_0 \):
(*) \( {^n b} < e < n \)
We also know that for \( b>\eta \), \( \exp_b^{\circ n}\to\infty \) quite fast, particularly for each \( b>\eta \) there is an \( n_0 \) such that for all \( n\ge n_0 \):
(**) \( {^n b} > n \).
Now we lead proof by contradiction, suppose that
\( \lim_{n\to\infty} b_n \neq \eta \) where \( {^n b_n} = n,\quad b_n > 1 \).
Then there must be a subsequence \( b_{m},\quad m\in M\subseteq\mathbb{N} \) and \( \epsilon>0 \) such that this subsequence stays always more than \( \eps \) apart from \( \eta \):
\( \left|b_{m}-\eta\right| \ge \epsilon \).
I.e. there is \( B_1<\eta \) and \( B_2>\eta \) such that
either \( b_m \le B_1 \) or \( b_m \ge B_2 \).
By (*) and (**) we have \( n_0 \) such that for all \( m\ge n_0 \):
\( {^m B_1}<m \) and \( {^m B_2}>m \).
As \( {^m x} \) is monotone increasing for \( x>1 \) we have also
\( {^m b_m}<m \) and \( {^m b_m}>m \).
This particularly means \( {^m b_m}\neq m \) and hence none of the \( b_m \) can be the self superroot, in contradiction to our assumption.
http://math.eretrandre.org/tetrationforu...73#pid4073
First it is easy to see that for \( 1<b<\eta \):
\( {^n b}=\exp_b^{\circ n}(1)\to a<e \) (\( a \) is the lower fixed point of \( b^x \))
Hence for \( n_0 > 3 \) we have for all \( n\ge n_0 \):
(*) \( {^n b} < e < n \)
We also know that for \( b>\eta \), \( \exp_b^{\circ n}\to\infty \) quite fast, particularly for each \( b>\eta \) there is an \( n_0 \) such that for all \( n\ge n_0 \):
(**) \( {^n b} > n \).
Now we lead proof by contradiction, suppose that
\( \lim_{n\to\infty} b_n \neq \eta \) where \( {^n b_n} = n,\quad b_n > 1 \).
Then there must be a subsequence \( b_{m},\quad m\in M\subseteq\mathbb{N} \) and \( \epsilon>0 \) such that this subsequence stays always more than \( \eps \) apart from \( \eta \):
\( \left|b_{m}-\eta\right| \ge \epsilon \).
I.e. there is \( B_1<\eta \) and \( B_2>\eta \) such that
either \( b_m \le B_1 \) or \( b_m \ge B_2 \).
By (*) and (**) we have \( n_0 \) such that for all \( m\ge n_0 \):
\( {^m B_1}<m \) and \( {^m B_2}>m \).
As \( {^m x} \) is monotone increasing for \( x>1 \) we have also
\( {^m b_m}<m \) and \( {^m b_m}>m \).
This particularly means \( {^m b_m}\neq m \) and hence none of the \( b_m \) can be the self superroot, in contradiction to our assumption.
I've been working on that one for while, ever since the 
