TITLE: A function that grows faster than any function in the sequence $e^x, e^{e^x}, e^{e^{e^x}}$...
QUESTION [10 upvotes]: 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.

REPLY [5 votes]: 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.