Isomorphism of group algebras of the dihedral group. I'm trying to solve the following problem.

Let $m,n \in \mathbb{N},  m|n$. Prove that $\mathbb{Z[D_{n}]}/ \langle R^{m} - 1 \rangle \sim \mathbb{Z[D_{m}]}$.

I'm trying to prove it using a hands on approach exhibiting the isomorphism between both rings without much luck. I can't think of any way of doing it. The hint given by the notes is that I should find a morphism and it's inverse, so I'm led to believe it shouldn't be that difficult but it is for now.
Any hint so that I can solve it will be more than helpful. Thanks a lot!
 A: Let $m | n$ be two positive integers. Consider the morphism
$$
\newcommand{\zd}[1]{\mathbb{Z}[\mathbb{D}_{#1}]}
\begin{align}
g : \zd{n}&  \to\zd{m} \\
 & 1 \mapsto 1 \\
 & r \mapsto \rho \\
 & s \mapsto \sigma
\end{align}
$$
with $r,s$ and $\rho,\sigma$ the generators of the corresponding dihedral groups. This mapping is clearly surjective, and we also have $\langle r^m-1 \rangle \subset \ker g$, since 
$$
g(a(r^m-1)) = g(a)(\rho^m-1) = g(a)(1-1) = 0.
$$
Via the first isomorphism theorem it would suffice to see the other inclusion, so that $\ker g = \langle r^m -1 \rangle$ and therefore $\zd{n}/\langle r^m-1\rangle \simeq \zd{m}$. 
Let $x = \sum_{s=1}^na_sr^s + bs\in \zd{n}$. Now, if
$$
0 = g(x) = \sum_{s=1}^na_s\rho^s + b\sigma\in \zd{m},
$$
then $b = 0$ and $\sum_{s=1}^na_s\rho^s = 0$. Noting $n = mk$, we get
$$
0 = \sum_{s=1}^na_s\rho^s = \sum_{i=1}^m\sum_{j=0}^{k-1}a_{mj+i}\rho^{mj+i} = \sum_{i=1}^m\left(\sum_{j=0}^{k-1}a_{mj+i}\right)\rho^{i} 
$$
and so 
$$
\sum_{j=0}^{k-1}a_{mj+i}  = 0 \quad (\forall i). \tag{$\star$}
$$
Hence, we have that 
$$
\begin{align}
x &= \sum_{i=1}^m\sum_{j=0}^{k-1}a_{mj+i}r^{mj+i} = \sum_{i=1}^mr^i\sum_{j=0}^{k-1}a_{mj+i}r^{mj} = \sum_{i=1}^mr^i\left[\sum_{j=1}^{k-1}a_{mj+i}r^{mj} + a_i\right]\\
& \stackrel{(\star)}{=} \sum_{i=1}^mr^i\left[\sum_{j=1}^{k-1}a_{mj+i}r^{mj} -\sum_{j=1}^{k-1}a_{mj+i}\right] = \sum_{i=1}^mr^i\sum_{j=1}^{k-1}a_{mj+i}(r^{mj}-1) \\
& = \sum_{j=1}^{k-1}(r^{mj}-1)\sum_{i=1}^mr^ia_{mj+i}.
\end{align}
$$
To see that $x \in \langle r^m -1 \rangle$, it suffices to see that each summand is in this ideal, and moreover we can reduce this to proving $(r^{mj}-1) \in \langle r^m -1\rangle$ for each $j \in \{1,\dots,k-1\}$. In effect, by the 'difference of powers' equality, 
$$
r^{mj}-1 = (r^m)^j - 1^j = (r^m-1)\sum_{l=0}^{j-1}(r^m)^l \in \langle r^m-1 \rangle.
$$
