Proving that $r^i \neq sr^j$ in $D_{2n}$

Question

I was trying to proof that the order of the dihedral group $D_{2n}$ is $2n$ given its presentation $$ D_{2n} = \langle r,s ~\vert~ r^n = s^2 = 1 ,~ rs = sr^{-1} \rangle $$ and I further assumed that the order of $r$ is precisely $n$ (someone please correct me if this is unnecessary assumption).
Anyway, I proved that there is at most $2n$ elements (by proving that all elements are of the form $s^ir^j$ for $0\le i<2$ and $0\le j<n$) and that no $2$ elements that have the same power of $(s)$ are equal (i.e. $s^ir^j \neq s^ir^k\hspace{0.2cm} \forall \hspace{0.1cm} j\neq k$), but regarding the third case (proving that $sr^i \neq r^j \hspace{0.2cm} \forall \hspace{0.1cm} 0\le i, j < n$), I am stuck at trying to prove that $s \neq r^k \hspace{0.2cm} \forall \hspace{0.1cm} 0\le k < n$.

My question is does the presentation of a group imply implicitly that any generator can't be an element of the cyclic group generated by another generator (i.e. $a_i \notin \langle a_j\rangle \hspace{0.2cm} \forall \hspace{0.1cm} a_i, a_j \in A $ and $ i \neq j$ where $A$ is the set of generators of the group). If not, can someone please tell me how to proceed from here.