A group $Q$ is called simple if $|Q|>1$ and the only normal subgroups of $Q$ are the trivial subgroups $\{e\}$ and $Q$.
Prove that for any finite group $G$ there exists a sequence of nested subgroups of $G$, $\{e\}=N_0\leq N_1\leq \cdots \leq N_n=G$ such that for each integer $i$ with $1\leq i\leq n$ we have $N_{i-1}\trianglelefteq N_i$ and the quotient group $N_i/N_{i-1}$ is simple.
My idea is to try induction on $|G|$. But I am having a hard time putting this into practice...
Evidently, $G/N$ is simple and nontrivial if and only if $N$ is a maximal proper normal subgroup. If $G\ne\{e\}$, the set of proper normal subgroups $G$ is ordered inductively by $\subseteq$, therefore it has maximal elements. If the group $G$ is finite, a recursively defined sequence $M_0\trianglerighteq M_1\trianglerighteq \ldots$ where each $M_i$ is a maximal proper normal subgroup of $M_{i-1}$ must terminate (say, with $M_k=\{e\}$), because their cardinalities are strictly decreasing. $N_i=M_{k-i}$ is a sequence such as those you want.