Proof that nilpotent groups are solvable

Proof

1. Since $G$ is nilpotent, it has a central series $\{e\} = N_0 \subseteq N_1 \subseteq \cdots \subseteq N_n = G$. Each $N_i$ is a normal subgroup of $G$, and $N_{i+1}/N_i \subseteq Z(G/N_i)$ for $i < n$.
2. Fix $i < n$. Since $N_i$ is normal in $G$, it is normal in the subgroup $N_{i+1}$. So $\{e\} = N_0 \trianglelefteq N_1 \trianglelefteq \cdots \trianglelefteq N_n = G$ is a subnormal series.
3. The factor $N_{i+1}/N_i$ is the set of cosets $xN_i$ with $x \in N_{i+1}$. Its product $(xN_i)(yN_i) = xyN_i$ is the product of the quotient group $G/N_i$.
4. By step 1, each such coset lies in the centre $Z(G/N_i)$. So for $x, y \in N_{i+1}$, we get $(xN_i)(yN_i) = (yN_i)(xN_i)$.
5. So each factor $N_{i+1}/N_i$ is abelian. By step 2, $G$ is solvable.