How to calculate $f(x)$ in $f(f(x)) = e^x$? How would I calculate the power series of $f(x)$ if $f(f(x)) = e^x$?  Is there a faster-converging method than power series for fractional iteration/functional square roots?
 A: There is a lot of material about this question here and in mathoverflow. There is also a "Tetration forum", where someone has implemented a version of tetration due to Hellmuth Kneser, see some forum entries there: http://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=8" also in citizendium there is an extensive article of Dmitri Kousznetzov who claims he has a usable interpretation (and implementation) see http://en.citizendium.org/wiki/Tetration
A: Look at this answer:
https://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx/44727#44727
In short, the analytic solution is
$$g^{[1/2]}(x)=\phi(x)=\sum_{m=0}^{\infty} \binom {1/2}m \sum_{k=0}^m\binom mk(-1)^{m-k}g^{[k]}(x)$$
$$g^{[1/2]}(x)=\lim_{n\to\infty}\binom {1/2}n\sum_{k=0}^n\frac{1/2-n}{1/2-k}\binom nk(-1)^{n-k}g^{[k]}(x)$$
$$g^{[1/2]}(x)=\lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k g^{[k]}(x)}{(1/2-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(1/2-k) k!(n-k)!}}$$
Insert here $g(x)=a^x$ The same way you can find not only square iterative root but iterative root of any order.
Unfortunately this does not converge for $g(x)=a^x$ where $a > e^{1/e}$. 
Here is a graphic for iterative root of $g(x)=(\sqrt{2})^x$

The question becomes more difficult when speaking about the base $a>e^{1/e}$. But in this case the solution can also be constructed, see this article.
A: Here's the proof of a theorem due to Thron (1956), extracted from a article of Laurent Bonavero (available at his webpage). 
Theorem. There is no entire function $f$ (that is $f:\mathbb C \to \mathbb C$ holomorphic)  such that $\exp = f \circ f$.
Proof. If such a function $f$ exists, then $f(\mathbb C)= \mathbb C^*$. Indeed, $f(\mathbb C) \supset \exp(\mathbb C) = \mathbb C^*$, but $0$ can't be in the image of $f$: if $f(x)=0$, then as $x \neq 0$, there exists $y$ such that $x=f(y)$ so that $\exp(y)=0$, absurd.
Therefore $f$ can be lifted by the exponential, $f=\exp \circ g \,$ for $g$ entire. So $\exp = \exp(g \circ f)$, and there must exist a constant $C$ such that $g \circ f(z)=z+C$ for all $z\in \mathbb C$. It follows that $f$ is injective, so $\exp$ should be injective too, which is absurd!
