what's the real value of $\sqrt2^{\sqrt2^{\large \sqrt2^{\sqrt2^{\unicode{x22F0}}}}}$? [duplicate]

Finding the value of an infinitely iterated function can be tricky. Usually, if $z$ is a fixed point of a function $f$ (that is, if $f(z)=z$), then it often happens that $$\lim_{n\to\infty} f^n(x)=z, \forall x$$ ...however, things tend to get a little bit complicated when $f$ has more than one fixed point, which is the case with your function $$f(x)=\sqrt 2^x$$ that has fixed points $$f(2)=2$$ $$f(4)=4$$ In this case, the value of $f^\infty(x)$ depends on $x$, which doesn't really make sense given the way you wrote it. Basically this means that its value depends on the number "on top" of your power tower. For example, if you start with $4$, and observe the sequence $$a_0=4$$ $$a_{n+1}=\sqrt 2^{a^n}$$ then each $a_n=4$, and so $$a_{\infty}=\lim_{n\to\infty}a_n=\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{\cdots}}}}=4$$ However, if you take $a_0=1$, $$a_{\infty}=\lim_{n\to\infty}a_n=\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{\cdots}}}}=2$$ However, things are a little bit less complicated in this case, since $x=2$ turns out to be something called an "attracting fixed point". Basically, we end up with $$f^\infty\left(x\right) = \left\{ \begin{array}{lr} 2 & : x \lt 4\\ 4 & : x = 4\\ \infty & : x \gt 4\\ \end{array} \right.\\$$