Finite Group has Composition Series/Proof 2

Proof
Let $G$ be a finite group whose identity is $e_G$. We shall use induction on $|G|$. If $G$ is trivial ($|G|=1$), then its composition series is
 * $G=\{e_G\}$.

Suppose $G$ has a composition series if $|G|<n$, then it suffices to construct a composition series for $G$ with order $n$. If $G$ is simple, then its composition series is
 * $G\supsetneq \{e_G\}$.

Otherwise, $G$ has one or more proper non-trivial normal subgroup. Let $S$ denote the set of all non-trivial normal proper subgroup of $G$. Notice that $S$ is non-empty, ordered by inclusion, and $G$ is an upper bound of every chain in $S$, then by Zorn's lemma, $S$ has at least one maximal element, denoted $H$. Since $H\in S$, $H$ is a non-trivial normal proper subgroup of $G$, in particular, $|H|<|G|$, so by the induction hypothesis, $H$ has a composition series. Now we have
 * $G\supsetneq H\supsetneq H_1\supsetneq \cdots \supsetneq H_n\supsetneq \{e_G\}$.

In order to show that it is indeed a composition series of $G$, it suffices to check that $G/H$ is simple, which is a direct consequence of the maximal property of $H$, so we are done.