Quotient of nilpotent group is nilpotent

I cleaned up the proof so that it no longer uses the somewhat unprofessional "coset multiplication" argument. I'm happy with it now, so I'll post it as an answer in case it is of interest to anyone.

Recall that the goal is to prove the following:

If $G$ is nilpotent and $N \lhd G$, then $G/N$ is nilpotent.

First, we prove a simple lemma:

---

Lemma: If $\theta : G \to H$ is an epimorphism, then $\theta(Z(G)) \leq Z(H)$.

Proof: Let $z \in Z(G)$ and $h \in H$. Then $h = \theta(g)$ for some $g \in G$, and $\theta(z)h = \theta(zg) = \theta(gz) = h\theta(z)$. Therefore, $\theta(z) \in Z(H)$.

---

Now we prove the following proposition:

Proposition: If $G$ is nilpotent and $\phi : G \to H$ is an epimorphism, then $H$ is nilpotent. (Note: Applying this proposition to the canonical epimorphism $G \to G/N$ will yield the desired result.)

Proof: Since $G$ is nilpotent, we have $N_i \lhd G$ and $1 = N_0 \leq N_1 \leq \cdots \leq N_r = G$, with $N_i / N_{i-1} \leq Z(G / N_{i-1})$.

Define $M_i = \phi(N_i)$. We claim that $1 = M_0 \leq M_1 \leq \cdots \leq M_r = H$ is a central series for $H$.

First, we observe that $M_0 = \phi(N_0) = \phi(1) = 1$ and $M_r = \phi(N_r) = \phi(G) = H$. Also, $M_i \lhd H$ by the correspondence theorem, since $N_i \lhd G$.

Now define $\theta : G / N_{i-1} \to H / M_{i-1}$ by $\theta(gN_{i-1}) = \phi(g)M_{i-1}$. It is routine to verify that $\theta$ is a well-defined epimorphism. We can therefore apply the lemma to conclude that $\theta(Z(G/N_{i-1})) \leq Z(H / M_{i-1})$.

Note also that $\theta(N_i / N_{i-1}) = \phi(N_i) / M_{i-1} = M_i / M_{i-1}$.

Since $G$ is nilpotent, we have $N_i / N_{i-1} \leq Z(G/N_{i-1})$. Consequently, $M_i / M_{i-1} = \theta(N_i / N_{i-1}) \leq \theta(Z(G/N_{i-1})) \leq Z(H / M_{i-1})$, as required.

Your argument is correct. A simpler and better way is as follows: $$N_i/N_{i-1} \leq Z(G/N_{i-1}) \Longleftrightarrow N_i/N_{i-1} \mbox{ commutes with } G/N_{i-1} \Longleftrightarrow [N_i/N_{i-1}, G/N_{i-1}]=1 \Longleftrightarrow [N_i,G]\leq N_{i-1} $$ By homomorphism $\phi$, we have $\phi[x,y]=[\phi(x),\phi(y)]$. Hence, $$ [M_i, H]=[\phi(N_i), \phi(G)]=\phi[N_i,G]\leq \phi(N_{i-1})=M_{i-1}.$$ This implies $H$ is nilpotent with central series $\{M_i\}_i$.