Why $rs=sr^{-1}$ holds with easy computation? 
Suppose $r,s\in S_n$ such that $r=(1\,2\,\ldots\,n)$ and $s=(2\,n)(3\,\,n-1)\cdots(k\,\,\,n+2-k)$ where $k$ is such that $n=2k$ if $n$ even and $n=2k-1$ if $n$ odd. I have to show that $rs=sr^{-1}$, but I am told that showing $rs(1)=2=sr^{-1}(1)$ and $rs(2)=1=sr^{-1}(2)$ is enough to establish $rs=sr^{-1}$.

Why we don't need to check for other naturals $3,\dots,n$?
 A: It's the fact that $r$ and $s$ are permutations of an $n$-gon, hence preserve adjacency, that allows us to worry about the effect of a permutation on two vertices only.
The idea is that any symmetry of the $n$-gon must send neighboring vertices to neighboring vertices. If the vertices are labeled cyclically, from $1$ to $n$, then they remain cyclically labeled after applying a symmetry of the $n$-gon. So, once you know $\pi(1)$ and $\pi(2)$, then the next vertex must be $\pi(3)$, and the one after it $\pi(4)$, and so on (otherwise, you haven't sent neighboring vertices to neighboring vertices).
As a concrete example, suppose $n = 7$, and $\pi(1) = 5$. Then, we only have two choices for $\pi(2)$: It must be something adjacent to $\pi(1) = 5$, hence can only be $4$ or $6$. If we choose $\pi(2) = 4$, then we must have the permutation
$$\pmatrix{1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 5 & 4 & 3 & 2 & 1 & 7 & 6}$$ (geometrically, a reflection across the line through vertex $3$)

If instead we choose $\pi(2) = 6$, then we're forced to choose $\pi(3) = 7$, etc, for the permutation
$$\pmatrix{1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 5 & 6 & 7 & 1 & 2 & 3 & 4}$$ 
(geometrically, a rotation of $3/7$-of-a-turn counter clockwise)

