The $9$th tetration of $-\sqrt2$ Let $^na$ denote the $n$th tetration of $a$, so that $^0a=1$ and
$$^{n+1}a=a^{^na}$$
for $n=0,1,\dots$. (For complex $x$ and $y$, here we use the definition $x^y:=e^{y\ln x}$, where $\ln$ is the principal branch of the logarithm.)
It appears that $^9(-\sqrt2)$ is very close to $1$, but not exactly $1$ — so that the sequence $\big(^n(-\sqrt2)\big)$ is almost(?) periodic:
$$^9(-\sqrt2)-1\approx(4.99+1.51\, i)\times10^{-45}.$$
Is this a mere coincidence or is there an explanation for this?
 A: This is not a huge coincidence: the idea is that the sequence $a_n={}^{n}(-\sqrt{2})$ has small norm until $n=6$, then it gets out of hand for $n=7$ ($a_7\sim-33+29i$), so that $a_8=e^{a_7\ln(-\sqrt{2})}$ is almost $0$ and $a_9$ is almost $1$.
In general when sequences of this type get out of hand (increase exponentially in norm), it seems likely that at some point you will raise $e$ to a number with a big negative real part so a similar phenomenon will happen.
In fact, as you say the sequence is almost periodic, in the sense that $a_{9n}$ converges to some point close to $1$ when $n$ goes to infinity. Let $k=\ln(-\sqrt{2}):=\pi i+\frac{1}{2}\ln(2)$ and let $f(x)=e^{kx}$, so that $a_{n+1}=f(a_n)$. Then $f'(x)=kf(x)$, and by the chain rule we get that $(f^n)'(x)=k^n\prod_{i=1}^nf^i(x)$.
Now let $g(x)=f^9(x)$. We know that $|g(1)-1|<10^{-40}$, so by the proof of the Banach fixed point theorem, to check that $g^n(1)$ converges to some point close to $1$ it is enough with the following proposition:
Proposition: If $x\in B(1,10^{-30})$, then $|g'(x)|<\frac{1}{2}$.
To prove it, note that $|g'(x)|=|k|^9\lvert\prod_{i=1}^9f^i(x)\rvert\leq10^6\prod_{i=1}^9|f^i(x)|$, so we just have to estimate $|f^i(x)|$. It is also easy to check that $|f^i(x)|<1.49$ for $i=1,\dots,6$. Also, for any $x$ with $|x|<1.5$ we have $|f'(x)|=|kf(x)|\leq4|f(x)|\leq4e^{3.5\cdot1.5}<1000$.
This implies by induction that $|f^i(x)-f^i(1)|\leq1000|f^{i-1}(x)-f^{i-1}(1)|$ for $i\leq6$, so $|f^6(x)-f^6(1)|\leq10^{-15}$, so $|f^7(x)-f^7(1)|<10^{-10}$ and $|f^8(x)|=|e^{kf^7(x)}|\sim |e^{-103-95i}|<|10^{-40}|$. So, finally, $|g'(x)|\leq10^6\prod_{i=1}^9|f^i(x)|\leq10^6\cdot1.49^7\cdot100\cdot10^{-40}\leq\frac{1}{2}$, as we wanted.
