The abstract description of the $n$th dihedral group $\mathcal{D}_n$. 
Every element of $\mathcal D_n$ is of the form $r^is^j$ for $\,0\le i<n,\,0\le s<2\,$, where $\mathcal D_n=\langle\{r,s\}\rangle$ with $|r|=n,|s|=2$ and $sr=r^{n-1}s.$

I want to prove the above written statement. But I am not able to form a formal proof of it.
We have: $\mathcal D_n=\{1,r,r^2,\ldots,r^{n-1},s,rs,r^2s,\ldots,r^{n-1}s\}\;(r^0=1).$ Here the generator $r$ denote a rotation $R_1$ through an angle $2\pi/n$ in the counter clockwise direction and $s$ is a reflection. I know that the rotations of $\mathcal D_n$ form the cyclic group $\langle r\rangle$ of order $n$ and so I can understand the first $n$ listing of $\mathcal D_n$ but I don't understand why each of the $n$ reflection elements can be expressed as $r^is$ for $0\le i<n.$
(It is known to me that a reflection followed by a rotation (or vice versa) is always a reflection. So $r^is$ indeed represents a reflection.)
Please help me to prove the statement mentioned at the very beginning. I want to know mainly about how we can deduce the form (viz., $r^is$) of the second half entries of $\mathcal D_n.$ Please give me some insights. Thanks in advance.
 A: Hint: show that all there are $n$ distinct reflections of the form $r^ks$.
A: I will start by noting one thing that you are not being asked to prove:

If $r^is^j$ and $r^as^b$ are two elements of $\mathcal{D}_n$, with $0\leq i,a\lt n$ and $0\leq j,b\lt 2$, and $r^is^j=r^as^b$, then $i=a$ and $j=b$.

That is: you are not being asked to prove that each element of $\mathcal{D}_n$ can be written uniquely in the given form; you are just being asked to prove that every element of $\mathcal{D}_n$ can be written, in that form, in at least one way.
The idea is that we can use the relation $sr=r^{n-1}s$ to "move" any $s$ that is to the left of an $r$ past it, so that we can "push" all the factors of $s$ to the right, all the factors of $r$ to the left, and then reduce using $|r|=n$ and $|s|=2$.
For example, say we had the expression $s^{-1}r^2sr^{-3}$, which is an element of $\langle r,s\rangle$. We can first change $s^{-1}$ to $s$, since $|s|=2$; then we can replace $r^{-3}$ with $r^{n-3}$, using $|r|=n$. And finally, we can use the relation $sr=r^{n-1}s$ repeatedly (note that $sr^k = r^{k(n-1)}s$) to get
$$\begin{align*}
sr^2sr^{n-3} &= sr^2r^{(n-3)(n-1)}s\\
&= r^{2(n-1)}sr^{(n-3)(n-1)}s\\
&= r^{2(n-1)}r^{(n-3)(n-1)^2}ss\\
&= r^{(n-1)(2+(n-3)(n-1))}s^2.
\end{align*}$$
Finally, we can reduce the exponent of $r$ modulo $n$ and the exponent of $s$ modulo $2$ to obtain an expression for this element in the desired form.
That's what we'd do "by hand". To prove it, perhaps the obvious thing to do is to use induction of some kind. Maybe induction on the number of times that $s$ appears in an expression of an element of $\mathcal{D}_n$.
So start with an arbitrary element of $\langle r,s\rangle$. This will be a product of powers of $r$ and $s$. Use $|r|=n$ and $|s|=2$ to replace any negative exponents with positive ones. So we know we can express every element of $\mathcal{D}_n$ as a product of positive powers of $r$ and $s$. Then you can try doing induction of the number of times that $s$ appears, to show you can always rewrite any such expression in the desired form.
