Why bother showing $S^{1}$ covers itself? I've just been introduced to covering spaces, and one of the examples I've been shown is that $p: S^{1} \to S^{1}$, $p(z)=z^{n}$ is a covering map for every $n$.  
My question is: why would you care? It's trivial that $S^{1}$ covers itself with the identity - why would you try and do this in many ways? Is it advantageous to have different ways to cover one space by another? 
 A: As an addition to Najib's answer, I would like to provide a nice fact which follows from the fact that $S^1$ covers itself. 
First of all, to be honest to myself, I will point out that showing $S^1$ covers itself just for the sake of examples is enough a reason.
Now, secondly, as a simple application, we have that, if a space $X$ is a finite CW-complex and we have a covering map $p: Y \to X$ with finitely many ($k$) sheets, then the Euler characteristics of the spaces are related by $\chi(Y)=k\chi(X)$. This follows from computing the Euler characteristic of a CW-complex, and the fact that the covering inherits the attaching maps from below by lifting, since the disk is simply-connected.
Now, since $S^1$ covers itself with a map of degree two ($z^2$), we must have that $\chi(S^1)=2\chi(S^1)$. Hence, $\chi(S^1)$ must be $0$.
Of course, this is rather trivial since the CW-complex structure of a circle is rather trivial. However, it also follows that the torus $\mathbb{T} ^n$ must have Euler characteristic $0$, since the product of covering maps is a covering map, and thus we have a covering map $\mathbb{T}^n \to \mathbb{T}^n$ of degree $2^n$ by making the product of $z^2 \times \cdots \times z^2$. And this is a less trivial fact.
A: It is interesting because these coverings have different degrees. They correspond to different subgroups of the fundamental group of $S^1$: the map $z \mapsto z^n$ corresponds to $n\mathbb{Z} \subset \pi_1(S^1) = \mathbb{Z}$. This shows that for different values of $n$, these coverings are not isomorphic, i.e. there is no homeomorphism $f : S^1 \to S^1$ such that $f(z)^n = z^m$ for nonnegative integers $n \neq m$.
And in fact you can show that, together with the universal cover $\mathbb{R} \to S^1$, these are all the connected coverings up to isomorphism. In other words, as soon as you have a connected cover $X \to S^1$, then either $X \cong \mathbb{R}$ and the covering is isomorphic to the exponential map, or $X \cong S^1$ and there exists a unique $n$ such that the covering is isomorphic to $z \mapsto z^n$. If such a classification result is not interesting, I don't know what is!
IMO it's a bit like asking

We already have a group (e.g. $\mathbb{Z}/4\mathbb{Z}$) with four elements, why do we care that we can construct another one ($\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$)?

Well these two groups are fundamentally different: they're not isomorphic. And you can show that up to isomorphic, these are the only two groups with four elements. Well, it's the same thing with the coverings of the circle: the power $n$ coverings are all different (not isomorphic), and together with $\mathbb{R} \to S^1$ you've completely exhausted all the possible coverings.
If I can phrase it yet another way: in a covering $E \to X$, the space $E$ is not all that matters; the map $E \to X$ matters too, in the same way that giving a group $G$ is not just giving a set, it's also giving a multiplication map. Sometimes you can take the same set but give it different multiplication maps that yield non-isomorphic groups.
A: You can view it as a toy example of classifying covering spaces of a given space. Every introduction begins with toy examples to understand basic concepts. In introductory group theory, subgroups of cyclic groups are classified for instance. There is a reason I use that analogy too: coverings of a space correspond to (conjugacy classes of) subgroups of a group, and $\pi_1(\mathbb{S}^1)\cong\mathbb{Z}$. Indeed, while that isomorphism is intuitively obvious, it is often proved using covering space theory on $\mathbb{S}^1$!
Pick  a handful of spaces. Can you identify all the coverings of those spaces without doing any hard work? Probably not, if you picked cool spaces. So perhaps the initial examples of coverings that you can get a handle on will include spaces that aren't very interesting to you. 
Keep in mind, by the way, that there are infinitely many distinct coverings of the circle. It's not very common for one space to cover another in multiple ways, but covering space theory cares about the covering map itself, not just the space upstairs, which is why classification here means figuring out all of the ways the circle can be covered. That it can only be finitely covered by itself and no other type of space (well, plus disjoint copies of itself) is just how the cookie crumbles.
To get a handle on understanding why the covering is more than just the space upstairs itself, consider covering spaces as a special case of fiber bundles with discrete fibers for a moment. A fiber bundle, denoted $F\to E\to B$, is comprised of the following data: a total space $E$, a base space $B$, and a "projection" $E\to B$ in which every fiber is a copy of $F$. Speaking geometrically, $E$ is a bunch of $F$s arranged in the shape of $B$. The trivial fiber bundle is $F\times B$. For instance, the direct product $[0,1]\times\mathbb{S}^1$ is a bunch of $[0,1]$s arranged in the shape of a $\mathbb{S}^1$. You can call that a cylinder, or better a wristband. But there is distinct fiber bundle $[0,1]\to M\to \mathbb{S}^1$, the Mobius band, which is also a bunch of $[0,1]$s arranged in the pattern of a $\mathbb{S}^1$!
The fact that coverings of $\mathbb{S}^1$ are all disjoint copies of itself can be visualized as a bunch of rubber bands, each one "wrapped around" an imaginary center a number of times. If we take a marker and mark a point on each wrapped band, then unwrap them we get a bunch of circles each with some number of marked "spokes" on them. Visually, this is the same as a cycle type and hence $n$-coverings of $\mathbb{S}^1$ correspond to conjugacy classes of the symmetric group $S_n$!
One can go further and classify principal $G$-bundles on $\mathbb{S}^1$ where $G$ is a finite group. (Such a thing is a covering space of $\mathbb{S}^1$ with $G$ acting by covering automorphisms in such a way that every fiber is acted on "regularly" by $G$, i.e. exactly like $G$ acts on itself.) Then one can understand the category of principal $G$-bundles over $\mathbb{S}^1$ with automorphisms: it's the action groupoid associated to $G$ acting on itself by conjugation! Since $\mathbb{S}^1$ is the only compact one-dimensional manifold, this gives the Dijkgraaf-Witten theory for cooking up 2D topological quantum field theories with finite gauge group $G$. TQFTs, axiomatized by Atiyah, are a way of viewing Feynman path integrals through the lens of category theory. Moral of the story: even toys can have applications.
