Denote by $D^iG$ the $i$th term in the lower central series of G, i.e. $D^0G = \{1\}$, $D^i=[G, D^{i-1}G]$. The claim is that if $N$ is normal in $G$, then $D^i(G/N) = (ND^i(G)/N)$.
I know we should proceed by induction, and we do thus:
For the base case, $i=0$, we have: $ D^0(G/N) = G/N \mbox{ and } (ND^0G)/N = (NG)/N$ But from the second isomorphism theorem, $(NG)/N \cong G/(G\cap N) = G/N$, and we have equality.
Now, assume that the claim holds for the $k$th term, and we will consider the $k+1$st term: $\begin{eqnarray*} D^{k+1}(G/N) &=& \left[ G/N, D^k(G/N) \right] \\ &=& [G/N, (ND^kG)/N] \\ &=& \langle [x, y] \mid x \in G/N, y \in (ND^kG)/N \rangle \\ &=& \langle xyx^{-1}y^{-1} \mid x \in G/N, y\in (ND^kG)/N \rangle \\ &=& \langle xyx^{-1}y^{-1}N \mid x \in G, y\in ND^kG \rangle \\ &=& \langle [x,y]N \mid x \in G, y\in ND^kG \rangle \\ &=& [G, D^kG]N \\ &\subseteq& (ND^{k+1}G)/N \end{eqnarray*}$
This gives one direction of inclusion. The other direction seems similar. Is this the right approach? I feel like I've bungled the chain of equalities. In particular, I'm pretty sure I just pulled the $ND^{k+1}G$ (necessary so that so that $N$ is normal in $ND^{k+1}G$) product out of thin air.
Tips, advice, corrections? (I'm a noob here, so be gentle :-P )