If $[G:H]=n$, then $g^{n!}\in H$ for all $g\in G$. I have the following question: 


  Let $G$ be a group and let $H$ be a subgroup of finite index of $G$. Let $|G:H|=n$
  Then it holds: $g^{n!}\in H$ for all $g\in G$.


Why is this true?

I think, that's not very difficult, but I have no idea at the moment. 

Thanks!
Let $G$ act on the left cosets of $H$ by left multiplication,
$$g: aH\longmapsto gaH.$$
This gives an action of $G$ on a set of $n$ elements, hence induces a homomorphism $G\to S_n$. In particular, for every $g\in G$, $g^{n!}$ must lie in the kernel, since $|S_n|=n!$. But the kernel is contained in $H$: if $g$ lies in the kernel, then $gaH = aH$ for all $a\in G$, hence in particular $gH=H$, so $g\in H$. Thus, for every $g\in G$, $g^{n!}\in H$.
This is an easy question if you start by showing that $g^n \in H$. Consequently, $g$ raised to the power of any multiple of $n$ is also in H (since $[G:H] = n$). Just an idea.