Comparison between two tetrations For a given natural number n, what ist the least number m, such that
$$e \uparrow \uparrow m > \pi \uparrow \uparrow n$$
It seems that m = n + 1 is the desired number. Is this true for all n ?
 A: Use the equation $\ln(\ln(b^{b^z}))=z\ln(b)+\ln(\ln(b))$ to help calculate the limit of the iterated logarithms of the iterated exponentiation of $\pi$.  For $k_6$, $z=\pi\uparrow\uparrow 4$, which has more than a billion digits and one can ignore the $\ln(\ln(b))$ term.  Then the approximation for $k_6$ without this term, is exactly the same as the equation for $k_5$ with the $\ln(\ln(b))$ term.
$$k_n = \ln^{[on]}(\pi \uparrow\uparrow n) 
\\ \lim_{n=6 \to \infty} k_n = k_5 + O\frac{1}{\pi \uparrow \uparrow 4}
\\ k_2 \approx 1.2798985874699298018433854796
\\ k_3 \approx 1.3167952100708763660472735425
\\ k_4 \approx 1.3176613010072958359625235748
\\ k_5 \approx 1.3176613010072958359630869692  $$  
For each value of n, $k_n$ can be thought of the value for the "n+1" term concatenated on top of the $e\uparrow\uparrow n$ tetration stack to get equivalence .  For n=3, $e\uparrow e \uparrow e \uparrow k_3 = \pi \uparrow \pi \uparrow \pi$.  For $\pi$, since $k_n$ is < e, as n gets arbitrarily large, 
$$\pi \uparrow\uparrow n < e \uparrow \uparrow (n+1)$$
As a side note, there is an interesting base $b\approx 7.28550781987618684208203148323$.  For this base b, the limit of the iterated logarithms is exactly e. 
$$\lim_{n \to \infty}\ln^{[on]}(b \uparrow\uparrow n)=e  
\\ \therefore \lim_{n \to \infty} b \uparrow\uparrow n = e \uparrow \uparrow (n+1)$$
$\pi$ falls into the same category as all bases$>$e, and $<$ b, where the limit of the iterated logarithms is greater than 1, and less than e. 
