Using Euler's formula $e^{ix} = \cos x + i\sin x$ at $x=\pi/2$ we get $e^{\pi i/2} = i$ Hence the expression has the value $i$.
My problem is how can an infinite power tree of real numbers have an imaginary answer?
(I really think I made some mistake, I'd be happy if it be pointed out)
In the real number system, it is evident that the number $\to\infty$
If we reduce the equation to $z=e^{\frac{\pi}{2}z}$, then there is no solution possible in real number space. It is in the analytic continuation of the product log function this equation possesses a solution which is $\pm i$.
If the sequence $z_0 = 1, z_{n+1}= \exp \frac{\pi}{2} z_n$ converges to some $\zeta\in \mathbb{C}$, then it converges to a point with $\zeta = \exp \frac{\pi}{2} \zeta$, and $\zeta = i$ is a solution of the latter equation. (The expression in the title doesn't quite make sense without some initial condition like $z_0 = 1$). The sequence $(z_n)$ doesn't converge, though; we have $z_{n+1} > 2z_n$ for (e.g.) real $z_n \geq 1$.
A simpler example of the same phenomenon is the sequence $z_0 = 2, z_{n+1} = z_n^2 + z_n + 1$. In the notation above, we would have $\zeta = i$. The sequence $z_n$ doesn't converge to $i$, though; the $z_n$ are real with $z_{n+1} > 2^n$ for all $n$, and $z_n \to\infty$ monotonically as $n\to\infty$.
Whether you were asking this question in terms of $e^\pi$ or $e^{i\pi}$, either way, they both diverge. In this other question, the radius of convergence was found to be: $$ \left[\frac{1}{e^e},e^\frac{1}{e}\right] $$ First and foremost: $$ e^{i\pi}=-1 $$ which shows that $e^{i\pi}$ is out of range. Secondly: $$ e^\pi>e^\frac{1}{e} $$ so $e^\pi$ is also out of range. $$ \text{Therefore, (by using Knuth's up arrow notation) both: }\lim_{n\to\infty}\left(e^{i\pi}\right)\uparrow\uparrow n\text{ and }\lim_{n\to\infty}\left(e^\pi\right)\uparrow\uparrow n\text{ diverge.} $$
