A function that grows faster than any function in the sequence $e^x, e^{e^x}, e^{e^{e^x}}$... Is there a continuous function constructed by elementary functions, or by integral formula involved only elementary functions (like Gamma function) that grows faster than any $e^{e^{e...^x}}$ ($e$ appears $n$ times)?
I ask for the answer with a single formula. Gluing continuous function together is too trivial.
The function need not to be defined on whole $\mathbb{R}$, the domain $(a, \infty)$ is acceptable.
 A: a negative answer
No, there is no such function constructed in a "natural" way, if we construe that word propertly.
Perhaps consult the literature on "transseries" ...
For example, transseries in the sense here:
G. A. Edgar, "Transseries for Beginners". Real Analysis Exchange 35 (2010) 253-310 .
The set of transseries includes real elementary functions, is closed under indefinite integration, composition, and many other operations.  
But no transseries has growth rate beyond all $e^{\dots e^x}$. There is an integer "exponentiality" associated with each (large, positive) transseries; for example Exercise 4.10 in:
J. van der Hoeven, Transseries and Real Differential Algebra (LNM 1888) (Springer 2006)
A (large, positive) transseries with exponentiality $n+1$ grows faster than any transseries with exponentiality ${} \le n$.
And $e^{\dots e^x}$ with $n$ exponentiations has exponentiality $n$.
A: Here is a proof that there is no such function und elementary function.
Let $A$ be a collection of functions which grow slower than the collection $e^x, e^{e^x} , \dots $.
We know that the sum, product composition and integral of functions which grow slower than $e^x, e^{e^x} , \dots $ also grows slower that $e^x, e^{e^x} , \dots $.
Then the set of functions which are finite iterate sums, $\dots$ of functions in $A$ also grows slower that $e^x, e^{e^x} , \dots $
Now, depending on or definition of elementary function, if you take $A= \{1, x, \exp, \log, x^a \}$ you should be done.
