Finite Group has Composition Series

Theorem
Let $G$ be a finite group.

Then $G$ has a composition series.

Proof
Let $G$ be a finite group whose identity is $e$.

Either $G$ has a proper non-trivial normal subgroup or it does not.

If not, then:
 * $\left\{{e}\right\} \triangleleft G$

is the composition series for $G$.

Otherwise, $G$ has one or more proper non-trivial normal subgroup.

Of these, one or more will have a maximum order.

Select one of these and call it $G_1$.

Again, either $G_1$ has a proper non-trivial normal subgroup or it does not.

If not, then:
 * $\left\{{e}\right\} \triangleleft G_1 \triangleleft G$

is a composition series for $G$.

By the Jordan-Hölder Theorem, there can be no other composition series which is longer. As $G_1$ is a proper subgroup of $G$:
 * $\left|{G_1}\right| < \left|{G}\right|$

where $\left|{G}\right|$ denotes the order of $G$.

Again, if $G_1$ has one or more proper non-trivial normal subgroup, one or more will have a maximum order.

Select one of these and call it $G_2$.

Thus we form a normal series:
 * $\left\{{e}\right\} \triangleleft G_2 \triangleleft G_1 \triangleleft G$

The process can be repeated, and at each stage a normal subgroup is added to the series of a smaller order than the previous one.

This process can not continue infinitely.

Eventually a $G_n$ will be encountered which has no proper non-trivial normal subgroup.

Thus a composition series:
 * $\left\{{e}\right\} \triangleleft G_n \cdots \triangleleft G_2 \triangleleft G_1 \triangleleft G$

will be the result.