Even with little/no knowledge of the symmetric group of degree $n$, I think that Cayley theorem is sufficient to infer that such group must have a proper subgroup of order $n$, for any $n>2$.
I was wondering what about groups of order $n!$: do they always have a proper subgroup of order $n$? This is the case for $S_n$. This is also the case for $C_6$, which complete the survey for $n=3$, but I can't even approach $n=4$ to look for counterexamples.
Best Answer
A (very) partial answer, for some special values, concerning the solvable case. (Note that solvable groups tend to have more subgroups than others.)
Indeed, write (uniquely) $n!=u_nv_n$ where $v_n$ is coprime to $n$, and every prime divisor of $u_n$ divides $n$. So $u_6=2^4.3^2$, $u_{10}=2^8.5^2$, $u_{12}=2^{10}.3^5$, $u_{14}=2^{11}.7^2$, $u_{15}=3^6.15^3$. Since every finite solvable group has a Hall subgroup for every set of prime numbers, for every $n$,
For each prime $p$, we denote by $N_p$ a maximal $p$-elementary normal subgroup of $G$. Any (solvable) group of order $n\ge 2$ has a minimal nontrivial subgroup $N$, which is elementary abelian. In other words, for some prime divisor $p$ of $n$ we have $N_{p}\neq 1$.
Next we check the above statement for $n\in\{6,10,14,15\}$. In each case, $n=pq$ with $p\neq q$ primes, and $u_n=p^aq^b$. The idea is to consider the action of a $q$-Sylow on $N_p$ and of a $p$-Sylow on $N_q$.
$\bullet$ For $n=6$, $p^aq^b=2^4.3^2$. The automorphism group of $N_3$ has order dividing the order of $\mathrm{GL}_2(\mathbf{F}_3)$, which is $3(3^2-1)(3-1)=2^3.3$. This is not divisible by $2^4$. Hence for a given $2$-Sylow, some element of order $2$ centralizes $N_3$; hence if $N_3\neq 1$ there is an element of order $6$.
We have $N_2\simeq(\mathbf{Z}/2\mathbf{Z})^s$ for some $s\le 4$. If $s\le 3$, the automorphism group of $N_2$ has order dividing the order of $\mathrm{GL}_3(\mathbf{F}_2)$, which is $2^3(2^3-1)(2^2-1)(2-1)=3.(2^3.7)$, which is not divisible by $9$, and we argue as in the previous case to get an element of order $6$. If $a=4$, the automorphism group of $N_2$ is isomorphic to $\mathrm{GL}_4(\mathbf{F}_2)$, which has order $2^6(2^4-1)(2^3-1)(2^2-1)(2-1)=3^2\ell$ with $\ell$ coprime to $3$. Since $\mathrm{GL}_2(\mathbf{F}_2)$ has an element of order $3$, we obtain a $3$-Sylow of $\mathrm{GL}_4(\mathbf{F}_2)$ by $2\times 2$ block diagonal matrices. In particular, an element in a single block (trivial on the second block) centralizes some element of order $2$, and hence there is an element of order $6$.
$\bullet$ For $n=10$, $p^aq^b=2^85^2$. We immediately see that $N_5$ has automorphism group of order not divisible by $2^8$, so there is an element of order $10$ as soon as $N_5\neq 1$.
If $|N_2|\le 2^7$, its automorphism group has order not divisible by $5^2$, so there is an element of order $10$ if $1<|N_2|<2^8$.
If $|N_2|=8$, a 5-Sylow in $\mathrm{GL}_2(\mathbf{F}_2)$, as in the previous case $n=6$, has order $25$ and is given by a product of 5-Sylow in block copies of $\mathrm{GL}_4(\mathbf{F}_2)$. As in that case, we deduce the existence of an element of order $10$.
$\bullet$ For $n=15$, $p^aq^b=3^6.5^3$. Then $3^6$ does not divide $|\mathrm{GL}_3(\mathbf{F}_5)|$, so if $N_5\neq 1$ then there is an element of order $15$. Also $5^3$ does not divide $|\mathrm{GL}_6(\mathbf{F}_3)|$ so if $N_3\neq 1$ then there is an element of order $15$.
$\bullet$ $n=21$ has $p^aq^b=3^9.7^3$ works exactly as in the previous case.
$n=12,14$ can be approached in the same way but I didn't push to a conclusion. For instance, for $n=14$ (so $p^aq^b=2^{11}.7^2$), $N_7\neq 1$ easily implies the existence of an element of order $14$; when $N_2\neq 1$, I don't immediately reach a conclusion in the case when precisely $|N_2|=2^9$.
Edit: I can also conclude the solvable case for $n=14$.
Here $p^aq^b=2^{11}.7^2$. If $N_7\neq 1$, as already said, $2^{11}$ does not divide $|\mathrm{GL}_2(\mathbf{F}_7)|$ so there is an element of order $14$.
If $N_2\neq 1$, write $|N_2|=2^s$, with $1\le s\le 11$. Since $2^k$ equals $1$ modulo $7$ iff $3$ divides $k$, one sees that if $s$ is not divisible by $3$, then every $7$-Sylow in $\mathrm{GL}_s(\mathbf{F}_2)$ fixes a nonzero vector, and hence we deduce in this case the existence of an element of order $14$.
Also $49$ does not divide $|\mathrm{GL}_3(\mathbf{F}_2)|$ so if $s=3$ then we also produce an element of order $14$.
If $s=6$, a $7$-Sylow in $\mathrm{GL}_6(\mathbf{F}_2)$, as in the cases $n=6,10$, has order $49$ and appears in diagonals blocks $3\times 3$, and in particular it contains nontrivial elements with nontrivial fixed points, and again we produce elements of order $14$.
Finally let us consider $s=9$: then the $7$-Sylow in $\mathrm{GL}_9(\mathbf{F}_2)$ have order $7^3$, and decompose according to diagonals blocks $3\times 3$. To be the identity on the first block defines a subgroup of index $7$. Also the action of the $7$-Sylow of $G$ (of order 49) goes through a subgroup of index set. Hence these two subgroups have nontrivial intersection, and hence we again produce an element of order $14$.
Edit 2 (August 11):
The solvable case was done before. So we have to look at the possible non-abelian simple subquotients. Namely, we need the list of simple groups $S$ such that $|S|$ divides $10!$, and among them single out those for which $|S|^2$ divides $10!$.
There are $12$ such nonabelian simple groups of order dividing $10!=2^8.3^4.5^2.7$, namely,
$\mathrm{Alt}_5$ of order $60=2^2.3.5$, $|\mathrm{Out}|=2$,
$\mathrm{PSL}_2(7)$ of order $168=2^3.3.7$, $|\mathrm{Out}|=2$,
The last two have an element of order 10. For the first 10, $25$ does not divide the order of the automorphism group. We then argue as follows: first, mod out by a maximal normal subgroup of order coprime to $10$. Then the resulting quotient has order divisible by $2^8.5^2$, and has a minimal normal subgroup of order not coprime to $10$. The case of an elementary abelian $2$-group or $5$-group was already tackled. Otherwise, we have $S$ or $S^2$ as above. In case this is $S$ (not $S^2$), and assuming $S$ not $J_2/\mathrm{Alt}_{10}$ (already tackled), some element of order $5$ in the $5$-Sylow normalizes this normal subgroup (of even order), so there is an element of order $10$.
Finally, we have the case when $S^2$ is a subquotient (with kernel of order coprime to $10$). The examples for which $|S|^2$ divides $10!$ are the first five. For those of order divisible by $10$, $S^2$ has an element of order $10$ and we are done. For those of order coprime to $5$, the automorphism group of $S^2$ has order coprime to $5$, and hence we argue as in the case of $S$.
It was already checked by another user that every group of order $6!$ has a subgroup of order $6$. Still I get an element of order $6$, except possibly in a subgroup of index $2$ in $\mathrm{Aut}(\mathrm{Alt}_6)$ (there are three, including $S_6$ which has an element of order $6$).