Does it converge or does it not?

Relevant wiki: Maclaurin Series

Let $S$ be the given expression. Then

$\begin{aligned} e^{\pi i} & = 1 + \pi i + \frac {\pi^2 i^2}{2!} + \frac {\pi^3 i^3}{3!} + \frac {\pi^4 i^4}{4!} + \frac {\pi^5 i^3}{5!} + \frac {\pi^6 i^6}{6!} + \cdots & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = 1 + \pi i - \frac {\pi^2}{2!} - \frac {\pi^3 i}{3!} + \frac {\pi^4}{4!} + \frac {\pi^5 i}{5!} - \frac {\pi^6}{6!} + \cdots \\ & = 1 - S \\ \implies S & = 1-e^{\pi i} = 1 - (\cos \pi + i\sin \pi) = \boxed{2} & \small \color{#3D99F6} \text{By Euler's formula: }e^{\theta i} = \cos \theta + i\sin \theta \end{aligned}$

You will get the answer 2 by using the taylor series of e^x replacing the x with pi*i and then once again replacing it with -1. Although furthur simplification would be required.