Iterated logarithm to the $n - 1$ of the auto tetration n^^n I'm considering the sequence

*

*$0 = \log(1)$

*$\log\left(2^2\right)$

*$\log\left(\log\left(3^{3^3}\right)\right)$

*$\log\left(\log\left(\log\left(4^{4^{4^4}}\right)\right)\right)$
I tried solving this using Python and I get lots of $\log(x < 0)$ problems. I thought they were mostly fence post and recursion type errors, so I debugged my code but still the latest iteration has those problems for $n > 5$ or so.
I'm beginning to suspect that the iterated logs are too powerful, even when taken down a notch, and that I should do something to the argument to keep things finite. My first theory, on discovering a recursive solution, was that my sequence was increasing but I'm not at all sure anymore
Just to clarify: by iterated logarithm I mean $\log(\log(\log(... \text{Self tetration}, n^{n^{\dots}} )))$, is just $1$, $2^2$, $3^{3^3}$, ...
Iterated logs to the $n$ are provably overkill and iterated logs to the $n-2$ fail to tame the power tower.
 A: I'm assuming these are natural logarithms, although as it turns out the base of the logarithm doesn't matter much. Write $\log^k$ for the iterated (natural) logarithm and $n \upuparrows k$ for tetration. Note that with base-$n$ logarithms we clearly have
$$\log^k_n (n \upuparrows n) = (n \upuparrows (n-k)).$$
For $n \ge 3$ we have $\log n > 1$ and hence $\log_n x < \log x$, which gives
$$\log^{n-1}(n \upuparrows n) > \log \log_n^{n-2} (n \upuparrows n) = \log n^n = n \log n.$$
In particular there should in principle be no problems with taking logs of negative numbers, but I don't know what any particular piece of Python code might be doing to compute logs here.
But we can say more than this: it turns out that the base of practically every logarithm in the sequence except the outermost one doesn't matter, so the above is a pretty accurate approximation, which means
$$\boxed{ \log^{n-1} (n \upuparrows n) \approx \log n^n = n \log n }$$
so this sequence is increasing except for possibly the first few terms which can be checked by hand. To give an idea of how this works here's the $n = 4$ case. We get
$$\log 4^{4^{4^4}} = 4^{4^4} \log 4$$
$$\log \log 4^{4^{4^4}} = 4^4 \log 4 + \log \log 4.$$
Here $4^4 \log 4 \approx 355$ and $\log \log 4 \approx 0.326$ is much smaller in comparison so can be safely ignored when we take the next logarithm, hence
$$\log \log \log 4^{4^{4^4}} \approx 4 \log 4 + \log \log 4 \approx 4 \log 4.$$
In general we can argue by induction that $\log^k (n \upuparrows n) \approx (n \upuparrows (n-k)) \log n$. The error terms are quite tiny in comparison to how big the power towers are until we get to the outermost or maybe the second-to-outermost log. I can be more precise about this if you want.
