I know that showing this idea very basic(as shown in method 1), but I tried something to do with "odd,even" permutation.
My question is proof of Lemma 4 in Method 2. (Proof have wrong logic or not.)
$\star$Method 1: Using contrapositive proof method.
Theorem:
If a group $G$ is cyclic, then it is abelien.
Contrapositive:
If a group $G$ is not abelien, it is not cyclic.
Since $S_n$ is not abelien for $n>2$ it is not cyclic for $n>2$.$\Box$
$\star\star$Method 2: Using $3$ wellknown permutation group theorem and additional key lemma (lemma 4)
Theorem 1: Every permutation can be written by product of transpositions.
Proof: Let $f=\left (f_n \; f_{n-1}\; ... f_3 \;f_2\;f_1\right)$ and for arbitrary $n\in\mathbb N^+$ $f$ is equal to $(f_{2}\;f_{1})(f_{3}\;f_{1})(f_{4}\;f_{1})....(f_{n-1}\;f_{1})(f_{n}\;f_{1})=\left (f_n \; f_{n-1}\; ... f_3 \;f_2\;f_1\right)$$\Box$
Theorem 2: Every permutation is written either odd or even number of transpositions
Proof: Omitted, it can be found, it is a little bit long.$\Box$
Lemma 3: For $n>2$ every $S_n$ has even and odd permutations.
Proof: $S_3$ has $(123)$ which is even and $(12)$ which is odd. Since $\forall n>3$,$S_n$ consists $S_3$, $S_n$ has even and odd permutations. $\Box$
Lemma 4: Every power of a permutation is the same polarity. That is, $f$ be permutation, and $P(x)$ be polarity function that shows permutations odd or even. $P(f)=P(f^2)=....=P(f^{d-1})$ where $o(f)=d$ (order of $f$)
Proof: Using $o(fg)=lcm(o(f),o(g))$ rule, I prove only for odd permutation's powers, since the proof for even permutations is the same.
Assume that $f$ is odd permutation, that is order of $f$ is even. $o(f^2)=lcm(o(f),o(f))=o(f)$ that is order of $f^2$ is even and so permutation $f^2$ is odd permutation. $\Box$
Main Theorem: $S_n$ is not cyclic $\forall n>3$
Proof: Assume that $S_n$ is cyclic for some $n>3$ so there must be an element that generates $S_n$, let call this element as $g$ and by assumption $<g>=S_n$ but it is wrong because $g$ can only generate like its polarity(recall $P(x)$ in lemma 4) (that is, if $g$ is even permutation, $g$ can only generate even permutation but there is odds ass well (lemma 3), and for $g$ is odd, logic is the same.)$\Box$
Here is a proof using cycle decomposition:
Every permutation is the product of disjoint cycles.
Every cycle in $S_n$ has order at most $n$.
The order of a product of disjoint cycles is at most the lcm of the orders of the cycles.
Therefore, the order of every permutation in $S_n$ is at most $lcm(1,2,\dots,n) < n!$ because $lcm(1,2,3,4) < 4!$.
Unfortunately, this argument needs $n \ge 4$. The case $n=3$ is easy.
Here is a proof that works for $n \ge 3$: