We can define by recursion the Lower Central Series for a group $G$ as: \begin{align} \gamma_0(G)=G,\quad \gamma_1(G)=[G,G],\quad \gamma_n(G)=[G,\gamma_{n-1}(G)] \end{align} I want to find $\gamma_n(D_m)$, where $D_m\cong\left\langle a,b\;:\;a^m=b^2=1,\,ab=ba^{-1} \right\rangle$ is the dihedral group.
My attempt:
$\gamma_0(D_m)=D_m$,
\begin{align} \gamma_1(D_m)&=[D_m,D_m]=\left\langle [a,b] \right\rangle=\langle a\underbrace{ba^{-1}}_{=ab}b^{-1} \rangle=\left\langle aabb^{-1} \right\rangle\\&=\left\langle a^2 \right\rangle \cong \begin{cases} \text{if }m\text{ is even:}\quad\left\langle a^2\;:\; (a^2)^{m/2}=1 \right\rangle\cong C_{m/2}\\ \text{if }m\text{ is odd:}\quad\left\langle a\;:\;a^m=1\right\rangle\cong C_m \end{cases} \end{align}
\begin{align} \gamma_2(D_m)&=[D_m,\gamma_1(D_m)]=[D_m,[D_m,D_m]]\\&= \begin{cases} \langle [a,a^2]\rangle=\langle aa^2a^{-1}(a^2)^{-1}\rangle =\langle a^2a^{-2}\rangle=\langle 1\rangle\\ \langle[b,a^2]\rangle=\langle ba^2b^{-1}(a^2)^{-1}\rangle=\langle baa\underbrace{b^{-1}}_{b^{-1}=b}a^{-2}\rangle =\langle ba\underbrace{ab}_{=ba^{-1}}a^{-2}\rangle=\langle b\underbrace{ab}_{=ba^{-1}}a^{-1}a^{-2}\rangle=\langle \underbrace{b^2}_{=1}a^{-1}a^{-1}a^{-2}\rangle=\langle a^{-4}\rangle \end{cases} \end{align}
So we only consider $\langle a^{-4}\rangle\cong\langle a^{-4}\;:\;(a^{-4})^{m/4}=1\rangle\cong C_{m/4}$.
If I repeat the process for $\gamma_3(D_m)$, I get that $[a,a^4]=1$ and
\begin{align} [b,a^{-4}]&=ba^{-4}b^{-1}a^4=\underbrace{ba^{-1}}_{=ab}a^{-3}b^{-1}a^4 =a\underbrace{ba^{-1}}_{=ab}a^{-2}b^{-1}a^4=\ldots=a^4bb^{-1}a^4=a^4a^4=a^8 \end{align} So I have a pattern like "$\gamma_n(D_m)\cong\langle a^{2^n}\rangle$", but I don't know if that is correct, if I apply induction on $n$.