Can we conclude any information about the isomorphism classes of groups of order $n!$? We know that there a isomorphism class of symmetric group structure of order $n!$.
Can we conclude any other information about the isomorphism classes of groups of order $n!$ ?
 A: For $n \in \mathbb{Z}^+$, let $g(n)$ be the number of isomorphism classes of groups of order $n$.  I think that most sufficiently inquisitive students of group theory have asked themselves what can be said about this function.  For instance:

*

*For which $n$ is $g(n) = 1$?

This is equivalent to: for which $n$ is every group of order $n$ cyclic.  This is known: it is necessary and sufficient that $\operatorname{gcd}(n,\varphi(n)) = 1$.  I believe the result goes back to Burnside; a nice treatment can be found here.


*What are upper and lower bounds on $g(n)$?

By 1), $g(n) = 1$ infinitely often.  It is certainly also at least two infinitely often: in fact for any prime number $p$, $g(p^2) = 2$.  So there is no hope for a precise asymptotic formula for $g(n)$.  The sequence is discussed here: by its label, I guess it is a rather important integer sequence!
There are certainly some interesting results here: for instance, for $n \in \mathbb{Z}^+$, let $m(n)$ be the largest power to which any prime divides $n$ (so e.g. $m(n) = 1 \iff n$ is squarefree).  Then:

Theorem (Pyber): $g(n) \leq n^{2/27 m(n)^2 + O(m(n)^{3/2})}$.

It is known that $g$ grows very rapidly along prime powers.  In fact, if you fix an exponent $a$ and ask about $g(p^a)$ for a variable prime number $p$, you get some very interesting questions, which are for instance related to elliptic curve theory (!!).  For a nice discussion, see e.g. here.  In particular that reference gives:

Theorem (Newman-Seeley): $p^{2/27 n^3 - 6n^2} \leq g(p^n) \leq p^{2/27 n^3 + O(n^{5/2})}$.

Okay, you asked about $g(n!)$.  The answer is that it will be huge.  This follows from the following (rather crude: surely someone else can do better) multiplicative property of $g(n)$:
$g(mn) \geq \max g(m), g(n)$.
Indeed this follows from:

Theorem Krull-Remak-Schmidt: if $G \times H \cong G \times K$, then $H \cong K$.

Thus $g(n!)$ is at least as large as $g(2^{\sum_{n=1}^{\infty} \lfloor \frac{n}{2^i} \rfloor})$, since $2^{\sum_{n=1}^{\infty} \lfloor \frac{n}{2^i} \rfloor}$ is the largest power of $2$ dividing $n!$.
I'm a little too pressed right now to write out the details, but if you put that together with the lower bound in the Newman-Seeley Theorem, you should get a lower bound on $g(n!)$ which is both explicit and huge.
