If there are entire $G_k$s such that $f=\exp\circ\exp\circ\cdots \circ\exp\circ G_k$ ($k$ times), must $f$ be constant? I am a French guest and I hope that my English isn't too bad...
So here is my issue:
I consider an entire function $f$ which satisfies the following property for each complex number $z\in \mathbb{C}$:
$\forall ~ k \in \mathbb{N}^*$, there exists an entire function $G_k$ that satisfies
$$f(z)=\exp_k(G_k(z))$$
where $\exp_k$ denotes $\exp \circ \exp \circ ...\circ \exp$, $k$-times.
In other words, I can take (as many times as I wish) the $\log$ of my function $f$, and it will always give an entire function that doesn't vanish on $\mathbb{C}$.
Is $f$ a constant? (surely different from $0$...)
Thanks everyone!
(PS: This forum is so cool!)
 A: I must say that at first I was quite skeptical  that such function could exist, but it turned out that one can prove its existence.  I have written down the proof but I will give here only sketch it (you should not have any problems computing it for yourself, nice exercise) 
Let us define $b_1=2\pi \rm i m_1$ and $\displaystyle b_n=2\pi \rm i m_n+ \log b_{n-1}$ and 
$$f_n(z)=\exp_{n}\left[b_n+ \frac{z}{b_1b_2\ldots b_{n-1}}\right],$$
where $\exp_{n}[z]=\exp\circ\cdots\circ\exp(z)$ $n$-times. One must find a sequence of  integers $m_k>0$ such that  the sequence  $f_n(z)$ will converge on every compact set. With some more effort one can prove that limit function $g(z)=1+z+ \mathcal{O}(z^2)$ have the desired property. 
A: I think that the concept of 'order of an entire function' can help you to get an answer.
(See definition in W. Rudin : Exercise 2, Ch. 15 of ; or alternatively here : Wiki - Entire function )
Namely, your entire function $f$ is clearly of infinite order (well, at least for k>1), while constant functions -like polynomials- are entire functions of order zero ... 
NB: of course I'm not considering the trivial case where the $G_k$ are constant (and non-zero), in which case $f$ would be constant as well ...  
