I am trying to understand the proof of the following proposition:
A finite group $G$ is soluble $\iff G^{(k)}=\{e\}$ for some $k$, where $G^{(0)}=G$, $G^{(i+1)}=[G^i G^i]$ is the derived series.
Proof: Since $G$ is finite there is some $i$ such that $G^{(i)}=G^{(i+1)}$, and clearly $G^{(i)}=G^{(k)}\ \forall k\ge i$. So $\{e\}$ appears in the derived series $\iff G^{(i)}=\{e\}$.
Suppose that $G^{(i)}\ne\{e\}$ and let $G^{(i)}=H_0>H_1>...>H_k=\{e\}$ be a composition series for $G^{(i)}$. Since $\color{red}{G^{(i+1)}\nleq H_1}$ and since $G^{(i+1)}$ is the derived subgroup of $G^{(i)}$, it follows that $H_0/H_1$ is non-abelian. So $G^{(i)}$ is not soluble and hence $G$ is not soluble (since any subgroup of a finite soluble group must be soluble).
Now suppose that $G^{(i)}=\{e\}$. We know that if $X$ is a finite group and $N\lhd X$, then $X$ has a composition series which includes $N$. $\color{red}{\text{Using this fact repeatedly we get a composition}}$ $\color{red}{\text{series for } G \text{ which includes } G^{(j)} \text{ for all } j\le i}$. It is clear that the composition factors are abelian since $G^{(j)}/G^{(j+1)}$ is abelian for all $j$. So $G$ is soluble. $\quad\quad\square$
I highlighted the parts which I didn't understand. Would somebody be so kind and explain the details. :/
1) "Suppose that $G^{(i)}\ne\{e\}$ and let ... Since $\color{red}{G^{(i+1)}\nleq H_1}$" -- since $G^{(i)}=G^{(i+1)}$.
2) "$\color{red}{\text{Using this fact repeatedly we get a composition series for } G \text{ which includes } G^{(j)}}$ $\color{red} {\text{ for all } j\le i}$" -- Take a composition series for $G^{(i-1)}$. Then take the series for $G^{(i-2)}$ which includes $G^{(i-1)}$, but change the begin of this series from $e$ up to $G^{(i-1)}$ by the first series which you chose for $G^{(i-1)}$ etc.