Let $n\in\mathbb{N}$ and let $G$ be a group. If there exists a sequence of normal subgroups of $G$ $$G=G^{0}\supset G^1\supset\ldots\supset G^n=\{e\}$$ such that the groups $G^k/G^{k+1}$ are abelian, then there exists a normal abelian subgroup $A$ of $G$ such that $G/A$ is solvable of class $\leq n-1$.
I think the proof is by induction on $n$. I am stuck on the base case $n=1$: suppose $G/\{e\}$ is abelian. This means that $G/\{e\}$ is solvable of class $\leq 1$.
But I have to show that it is solvable of class $\leq n-1=1-1=0$: i.e. that $G/\{e\}=\{e\}$. But why should this follow? What is my mistake?
You don't have to show that $G$ is solvable of class $0$. There needs to be an abelian normal subgroup $A$ such that $G/A$ is solvable of class $0$. You can take $A=G$.