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No. If $|z|<1$ the limit is $0$, if $|z|>1$ the limit is $\infty$ or does not exist since $|z|^n\to\infty$. If $|z|=1$ then $z=e^{i\theta}$ for $\theta\in[0,2\pi)$ and $z^n=e^{in\theta}$. Unless $\theta=0$, i.e. $z=1$, the sequence will rotate the point on the unit circle by $\theta$ counterclockwise at each step, and there is no limit.


WLOG, let $y = f(x)$. You know that $$y =x^y$$ So $$\ln(y) =y\ln(x)$$ $$\frac{dy}{dx}*\frac{1}{y} = \frac{dy}{dx}*\ln(x) + \frac{y}{x}$$ $$\frac{dy}{dx} = \frac{y^2}{x-xy\ln(x)}$$


This function is known as infinite tetration. The sequence is undefined for $x\leq 0$ except at some rational points, so there is no point talking about the derivative there. Ditto for large $x$, where the sequence diverges to $\infty$. Rather surprisingly for $x>0$ the limit only exists for $x\in[e^{-e},e^{1/e}]$ as shown by Euler, see here and here. ...


It is not only true, but mentioned explicitly in the Wikipedia article on the subject. :-) In fact, it has been known since the time of Euler. What you forgot to add is that infinite tetration only converges for $x\in\Big[e^{-e}~,~\sqrt[e]e\Big]$.


I think I know what you're trying to say, but I'm also not sure if it is possible to say with any rigor. I think that you essentially want to essentially know whether "simplification" is in some sense "complete": so that if two recurrence relations which appear in different forms always give the same outputs given the same inputs, then they can both be ...


Any relation that produces this sequence can of course be 'ultimately' simplified to $a_{n+1}=2^{a_n}$ just because this is how consecutive elements in the sequence are related. You do not however specify which identities are 'allowed' in a simplification. There are analytic functions that vanish on all integers, $\sin(\pi x)$ for example, so ...

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