Lower Central Series Problem: $D^i(G/N) = (ND^i(G))/N$ 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 )
 A: Induction certainly works. (There are other ways of proving it if you know a bit more theory, but not necessary here). So your approach is a good one.
Some comments, then:


*

*You don't need the Second Isomorphism Theorem in the base case: $GN = G$, so $GN/N = G/N$ by simple set equality.

*You can save yourself a bit of work by noting that $ND^i(G) = D^i(G)N$ for all $i$; this because $N$ Is normal. That means that $yN\in ND^i(G)/N$ if and only if there exists $y'\in D^i(G)$ and $n\in N$ such that $yN=y'nN=y'N$; so you can assume $y\in D^i(G)$ directly, which makes some computations easier.

*It would be better if you don't switch between $x$ representing elements of the quotient and elements of the original group. So it would be better if your third line was 
$$ = \Bigl\langle [xN,yN] \Bigm| xN\in G/N, yN\in D^k(G/N)\Bigr\rangle$$
and then you explicitly used the induction hypothesis, by writing it as
$$ = \Bigl\langle [xN,yN]\Bigm| x\in G, y\in D^k(G)\Bigr\rangle$$
(already using the equality above).

*The other inclusion is actually easier: if you are assuming that $ND^kG/N = D^k(G/N)$, then look at a generator of $D^{k+1}G$: it is of the form $[x,y]$ with $x\in G$ and $y\in D^k(G)$. Therefore, $yN \in ND^k(G)/N = D^k(G/N)$, so $[x,y]N = [xN,yN]\in [G/N,D^k(G/N)] = D^{k+1}(G/N)$. Hence, $ND^{k+1}G/N\subseteq D^{k+1}(G/N)$. 

*Likewise, to show that $D^{k+1}(G/N)\subseteq ND^{k+1}G/N$, it suffices to show that every generator of $D^{k+1}(G/N)$ is the image of an element of $D^{k+1}G$, so just take $[xN,yN]\in D^{k+1}(G/N)$, with $xN\in G/N$ arbitrary and $yN\in D^{k}(G/N)$; then you can use the induction hypothesis directly on $yN$. 
