An important point is that a "pole" is actually the same thing as a removable singularity, if we think of our function as a map that takes values on the Riemann sphere (which is the complex plane with a point at $\infty$ added; the complex structure near $\infty$ comes from the map $z\mapsto 1/z$).
So a function that has a removable singularity or a pole at $z_0$ doesn't have a "real" singularity there at all; rather, we can extend the function to an analytic or meromorphic function in $z_0$. If we cannot extend the function in this way, the singularity is indeed "essential"; i.e., we cannot get rid of it. Thus the terminology is not one that is merely used for convenience or pedagogical purposes; rather, it is extremely natural.
As has already been mentioned, the magic of complex numbers results in many beautiful facts about essential singularities: functions with these singularities are very far from extending continuously.
The simplest of these facts is the Casorati-Weierstraß Theorem: The image of a neighborhood of an essential singularity is dense in the complex plane.
This is just a consequence of the removable singularities theorem. (If $f$ omitted a neighborhood of $a$, we could postcompose $f$ with a Möbius transformation that takes $a$ to infinity and see that the resulting function has a removable singularity.)
The most well-known result of this type is Picard's theorem which was already mentioned.
There are various beautiful strengthenings of Picard's theorem that arise from Nevanlinna theory, and Ahlfors's theory of covering surfaces.
So all essential singularities have some things in common, but on the other hand this should not lead us to believe that they are all the same. What they have in common is complicated behaviour, but they can be complicated in very different ways! Indeed, different transcendental entire functions (those that have an essential singularity at infinity; i.e. are not polynomials) can vary very much with respect to their behavior near infinity. Just for example, for some such functions, such as $z\mapsto e^z$, there exist curves tending to infinity on which the function is bounded, while for others this is not the case.
You don't!
The function in question has simple poles. The easiest way to see this is to note that $\exp(2n\pi i)=1$, so by continuity $e^z-1$ tends to zero as $z \rightarrow 2n\pi i$ and thus the reciprocal tends to infinity.
To show that the poles are simple, note that $e^z-1$ has a zero of degree $1$ at $2n\pi i$ so the reciprocal has a simple pole and so the residue is:
$$\frac{1}{\frac{d}{dz}(e^z-1)}=\frac{1}{e^{2n\pi i}}=1$$
Best Answer
$f$ is meromorphic in a domain $D \subset \Bbb C$ if there is a (possibly empty) set $P \subset D$ such that
Now assume that $f$ is not identically zero, and let $Z$ be the (possibly empty) set of the zeros of $f$. (We know from the identity principle that every point in $Z$ is isolated in $D$.)
The $h = f'/f$ is holomorphic in $D \setminus (P \cup Z)$.
At a $k$-fold pole of $f$ at $a \in P$ is $$ f(z) = (z-a)^{-k} g(z) $$ where $g$ is holomorphic and non-zero in a neighborhood of $z=a$. Therefore $$ h(z) = \frac{f'(z)}{f(z)} = \frac{-k}{z-a} + \frac{g'(z)}{g(z)} $$ in a neighborhood of $z=a$, so that $h$ has a simple pole at $z=a$.
Similarly, at a $l$-fold zero of $f$ at $b \in Z$ is $$ f(z) = (z-b)^{l} h(z) $$ where $h$ is holomorphic and non-zero in a neighborhood of $z=b$. Therefore $$ h(z) = \frac{f'(z)}{f(z)} = \frac{l}{z-b} + \frac{h'(z)}{h(z)} $$ in a neighborhood of $z=b$, so that $h$ again has a simple pole at $z=b$.
So the only singularities of $f'/f$ occur at the zeros and poles of $f$, and these singularities are poles.