If $G$ is an inﬁnite simple group then any proper subgroup of $G$ has inﬁnite index. 
If $G$ is an inﬁnite simple group then any proper subgroup of $G$ has inﬁnite index. 

This question's hint is use the $n!$-theorem but i dont understand how i use it for answer.
$n!$-theorem: Let $G$ be a group and $H$ be a subgroup of $G$ of finite index, say $|G:H|=n$. Then there is a normal subgroup N of $G$ such that $N\subseteq H$ and $G/N$ is isomorphic to a subgroup of $S_n$ and so $|G/N|$ divides $n!$. Indeed, ${\rm core}_G(H)$ is such a normal subgroup of $G$.
 A: Assume for contradiction that $G$ is an infinite simple group and has a finite index subgroup $H$ such that $|G:H|>1$.  Then, $\mathrm{Core}_G(H)$ is a finite index normal subgroup because the intersection of finite index subgroups is finite and by definition $\mathrm{Core}_G(H)$ is invariant under conjugation by $G$.  In particular, $G$ has a non-trivial proper normal subgroup and therefore cannot be simple.  Thus, $G$ does not have any proper finite index subgroups.
A: From a stylistic perspective, quoting a "theorem" here this is unnecessarily complicated and makes things harder to understand.  The proof of this statement relies on two basic facts; and these are facts that everyone who studies groups should know anyways.  1. There are only finitely many subgroups of a given finite index.  2. The intersection of two finite index subgroups has finite index.  I will take these for granted; each is an easy exercise.
Now given a finite index subgroup $H$, it has only finitely many conjugates by 1.  So intersect all of them.  This subgroup has finite index by 2.  It is clearly normal because conjugating the intersection amounts to intersecting the conjugates, but that just permutes the things we're intersecting and we end up intersecting the same finite set of subgroups.
