kth derivative of the nth tetration of x Suppose $f_n(x)$ denotes the $n$th tetration of $x$, e.g.
$f_4(x) = x^{x^{x^x}}$
Is there a relatively non-painful way to prove that the sequence of higher-derivative values
$f_n^{(k)} (1)$ for $n \geq 1$, 
is eventually constant? 
Edit:  Perhaps this would follow by showing that these derivatives are generally integers and by showing that the sequence of functions $f_n(x)$ converges uniformly in an interval around $1$ as $n \to \infty$?  But I don't know if proving these values are all integral is any easier than proving the sequences are eventually constant.
 A: We can prove the following by strong induction;
For all $N \in \mathbb{N}$, $\forall n \geq N$:
$$f_n^{(N)}(1)=f_N^{(N)}(1)$$
The key idea is the following

If $\forall k \leq N-1$, $f^{(k)}(1)=g^{(k)}(1)$ then:
  
  
*
  
*we have:
  $$\forall k \leq N-1, (\exp(f))^{(k)}(1)=(\exp(g))^{(k)}(1)$$
  
*with $\tilde{f}(x)=\ln(x) f(x)$, $\tilde{g}(x)=\ln(x) g(x)$ we have:
  $$\forall k \leq N, \tilde{f}^{(k)}(1)=\tilde{g}^{(k)}(1)$$

and is the results of a direct computation. All it remains is to use it with the recurrence formula:
$$f_{n+1}(x)=\exp(\ln(x) f_n(x))$$



*

*For $N=1$ we have:
$$f_{n+1}=x^{f_n(x)}$$
so:
$$f_{n+1}'(1)= \left(\frac{f_n(1)}{1}+f_n'(1) \ln(1) \right) 1^{f_n(1)}=1$$
so for all $n$:
$f_n'(1)=1=f_1'(1)$.


Remark: we could have started at $N=0$ but here we can see what will be the general method.


*

*Let $N>1$ and suppose that the result is true up to the rank $N-1$. Let $n \geq N$.


We have:
$$f_{n}(x)=\exp( \ln(x) f_{n-1}(x) )$$
But for all $k \leq N-1$, $f_{n-1}^{(k)}(1)=f_{N-1}^{(k)}(1)$ so for all $k \leq N$, $\tilde{f_{n-1}}^{(k)}(1)=\tilde{f_{N-1}}^{(k)}(1)$. We can then conclude that:
$$\forall k \leq N, f_n^{(k)}(1)=\exp(\tilde{f_{N-1}})^{(k)}(1)$$
which is independent of $n \geq N$.
