Understanding a proof that $z\mapsto z^n$ is a covering map from $S^1$ to $S^1$ 
Let $S^1 = \{z\in\mathbb C : |z|=1\}$. Define $p:S^1\to S^1$ by $p(z) = z^n$, $n\in\mathbb Z\setminus\{0\}$. Then $p$ is a covering map.

The proof in question states that the sheets of $S^1\setminus\{1\}$ are $n$ disjoint open arcs of angle $2\pi/n$ in the covering space $S^1$, and $p^{-1}(\{1\})$ contains the endpoints of all these arcs.
I get that $$p^{-1}(\{1\}) = \{e^{2\pi k i/n} : k=0,1,\ldots,n-1\}$$ and these arcs are between the points in this set. But what is meant by "sheets" and what is the significance that $p^{-1}(\{1\})$ contains their endpoints? I do not see how this shows that every point $z\in S^1$ has a neighborhood that is evenly covered by $p$. Is it that the slices of an open set $U\subset S^1$ are these (open) arcs?
Edit: I have an alternative "proof." For $z=e^{it}\in S^1$, choose the neighborhood $U= \{e^{it}:t-\pi i/|n| < t < t+\pi i/|n|\}$. Then it is claimed that the sheets over $U$ consist of $|n|$ disjoint open arcs of $S^1$, evenly spaced about the circle. Intuitively it makes sense, but how can I see it rigorously? Were this to be the case, then letting said arcs be $V_1,\ldots,V_n$, the restrictions $p\vert V_j :V_j\to U$ are homeomorphisms as both $V_j$ and $U$ are homeomorphic to $(0,1)$ and the map $p$ is invertible; namely $p^{-1}(z) = z^{1/n}$ which is clearly continuous.

Quoting Wikipedia: 

The homeomorphic copies in $C$
  of an evenly covered neighborhood $U$ are called the sheets over $U$ (where $p:C \to X$ is the covering map in question).

followed by some pictorial descriptions.
This answer gives a nice explanation why $z \to z^n$ is a covering map of $\Bbb S^1$.
