Every finite group is isomorphic to some permutation group.
Any group of order $n$ can be embedded into $S_n$.
(We say that group $G_1$ is embedded into $G_2$ if there is $f:G_1\to G_2$ that is homomorphism and injective.)
Now using this theorem we need to prove that there exists finitely many groups of order n.
My work.
Because every finite group of order $n$ can be embedded into $S_n$ and number of permutations is $n!$ so number of groups of order $n$ is $\leq \binom{n!}{n}$. Is this right? If yes how show it is always strictly less.
Sorry if this is very simple question.
The similar question says the same thing I am saying that there are at most $ \binom{n!}{n}$ possibilities. Question is why it can't be exactly $ \binom{n!}{n}$ but is strictly less than that.
Your reasoning isn't enough: why is it not the case that there are infinitely many groups of order $4$? You've said all such groups are isomorphic to permutation groups, but why can't there be a group of order $4$ isomorphic to $S_{100}$? You're using the symbol $n$ to denote at least two possibly-different numbers.
You're going to need some stronger property of the permutation groups; presumably your proof of that theorem actually gives you a lot more information than you've stated.