Let's twist Euge's proof into what I like to call "The Devil's proof".
Suppose the theorem were false. Then for some finite $G$, it does *not* have a composition series.
Let $n = \min\{|G|: G \text{ does not have a composition series}\}$.
If $G$ were simple, it would have the composition series $G_0 = G \supsetneq \{e\}$.
(in other words we pick $G$ to be a minimal counter-example).
Since $G$ is not simple, it has a normal nontrivial proper subgroup, say $N$.
Since $|N| < |G|$ (it is a proper subgroup of a finite group), we must have that $N$ has a composition series, say:
$N \supsetneq N_1 \supsetneq \cdots \supsetneq N_k = \{e\}$.
Since $N$ is nontrivial, and normal, we have:
$|G/N| = |G|/|N| < |G|/1 = |G|$, so $G/N$ has a composition series, say:
$H_0 = G/N \supsetneq H_1 \supsetneq \cdots \supsetneq H_{k'} = \{e_{G/N}\} = \{N\}$.
Let $\pi_N: G \to G/N$ be the canonical homomorphism $g \mapsto gN$.
Set $K_j = \pi_N^{-1}(H_j)$ for $j = 0,1,\dots,k'$ (the pre-image of the $j$-th subgroup in the composition series for $G/N)$.
It ought to be "obvious" that $K_j$ is a subgroup of $G$, for each $j$, but we can prove it, to be perfectly explicit.
Suppose $x,y \in K_j$. Note this means that $\pi_N(x)$ and $\pi_N(y)$ lie within $H_j$. Since $H_j$ is a subgroup of $G/N$ we have that:
$\pi_N(x)\pi_N(y) \in H_j$ ($H_j$ is closed under multiplication).
Since $\pi_N$ is a homomorphism, $\pi_N(x)\pi_N(y) = \pi_N(xy) \in H_j$.
But this means $\pi_N$ maps $xy$ into $H_j$, so $xy \in \pi_N^{-1}(H_j) = K_j$, so $K_j$ is closed under multiplication.
Since $G$ is finite, and $K_j$ is a subset of $G$ closed under multiplication, $K_j$ is a subgroup if it is non-empty.
But if $n \in N$, we have $\pi_N(n) = nN = N \in H_j$ (since every subgroup of $G/N$ contains the identity of $G/N$).
This shows that not only is $K_j$ non-empty, it also contains all of $N$ (as a subgroup), and since $N \lhd G$, it is true "all the more" that $N \lhd K_j$ (for any $j$).
Now we have: $K_j/K_{j+1} \cong (K_j/N)/(K_{j+1}/N)$. I claim $K_j/N = H_j$. For if $x \in K_j$, then $\pi_N(x) \in H_j$ (by the way we defined $K_j$), which is to say $xN \in H_j$. This shows that $K_j/N \subseteq H_j$.
On the other hand, if $yN \in H_j$, then $\pi_N(y) \in H_j$ (by the definition of $\pi_N$), whence $y \in \pi_N^{-1}(H_j) = K_j$, so that $yN \in K_j/N$. Since $H_j/H_{j+1}$ is simple (since we have a composition series for $H_0 = G/N$), it follows that $K_j/K_{j+1}$ is likewise simple, via the isomorphism above.
Thus we have:
$G = G_0 = K_0 \supsetneq K_1 \supsetneq \cdots \supsetneq K_{k'} = N \supsetneq N_1 \supsetneq N_k = \{e\}$
is a composition series for $G$, contradiction.
Thus no counterexample exists, and the theorem is true.
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This kind of proof is sort of "sneaky", one winds up wondering "what does induction have to do with this"? The pivotal step is in assuming the minimality of the order of $G$. Finite groups have orders which are natural numbers, and any non-empty set of natural numbers has a least element (this is called the *well-ordered* property of the natural numbers, and is equivalent to the principle of induction). If there are ANY finite groups that do not possesses composition series then the set of orders of such groups forms a non-empty subset of the natural numbers (there is a slight subtlety, here-the collection of all finite groups is "too big" to be a set, but the collection of all orders of those groups is indeed a set-the natural numbers (that this set is *all* of the natural numbers is evidenced by the existence of $\Bbb Z_n$, for any natural number $n$)).
Technically, Euge's proof is a "strong induction proof", and what I have written above is induction "turned inside-out".
