Let $S_{G:H}$ be the set of permutations of $G:H$. Consider $\phi:(G:N)\rightarrow S_{G:H}$ that sends $xN$ to $T_x$ (Where $T_x$ is the permutation of $G:H$ that sends $aH$ to $xaH$ ). It is easy to verify that $\phi$ is an injective group homomorphism. Thus, $G:N$ is isomorphic to a subgrop of $S_{G:H}$. Using Lagrange's theorem, we deduce that $|G:N|$ divides $|S_{G:H}|=n!$
A pity such a question will remain unanswered. Define a group action
$$X:=\left\{\;gH\;;\;g\in G\;\right\}\;,\;\;G\times X\to X\;,\;\;x(gH):=(xg)H$$
Prove the above indeed is an action of $\,G\,$ on $\,X\,$ , and thus we get the induced homomorphism $\,\phi:G\to \text{Sym}(X)\,$ . The kernel of this homomorphism, also known as the core of $\,H\,$ in $\,G\,$ , is characterized as the largest normal subgroup of $\,G\,$ contained in $\,H\,$ .
But since $\,G\,$ is simple we get then that $\,\ker\phi=1\,$ , and this means we can embed $\,G\,$ into $\,\text{Sym}(X)\,$ , and this means, by Lagrange's Theorem, that $\,|G|\,\mid\,[G:H]!\,$ ...
Best Answer
The action of $G$ on the cosets of $H$ gives a homomorphism $G\to S_n$. This must be injective, else the kernel would be a normal subgroup of $G$.
Now we have $n|H|=|G|$ and $|G|$ divides the order of $S_n\ldots$