Say we are given a function $$\Gamma(x)=f_1(x)^{f_2(x)^{\cdot^{\cdot^{f_n(x)}}}}$$
where $f_i,i\in[1;n]$, are continuous functions in their domains. Also assume that the function makes sense, e.g., if $f_n:\Bbb{R\to R},$ then $f_i>0, i \in[1;n-1]$.
Is it fair to say that $\Gamma(x)$ is then also continuous in its domain?
First, we note that for $x>0$ and $\alpha \in \Bbb{R}$ $$x^\alpha=e^{\alpha \ln x}.$$ If $\alpha >0$ and $\beta \in \Bbb{R}$, $$x^{\alpha^\beta}=e^{\alpha^\beta \ln x}=e^{e^{\beta\ln\alpha}\ln x}.$$
Similar pattern works for $\Gamma(x)$. I have already proven the continuity of exponential and logarithm functions. Furthermore, the result
$$\lim_{x\to \xi}g(x)=g\left(\lim_{x\to\xi}x\right)$$
holds if $g(x)$ is continuous at $x=\xi$.
Here I would very much would like to conclude that $\Gamma(x)$ is continuous in its domain. Or, is something missing?
(I am currently ignoring cases where $f_i<0$ and $f_{i+1}\in \Bbb{Q}$, should such an operation be valid.)
Bonus question
Sums are conveniently written with Greek capital 'sigma', or $\sum$. For products, the notation is capital 'pi', i.e, $\prod$. Is there a similar convention for stepwise exponentiation of exponents, as given above?
All you need to do is show $f,g$ continuous $\Rightarrow$ $f^g$ continuous, the general result follows by induction. The basic results follows from $f(x)^{g(x)}=e^{g(x)\ln f(x)}$.