If $G$ acts $k$-transitive and $k > 5$ and $G$ is neither alternating nor symmetric, then $(n-k)! \ge 2n$ The following is an exercise from D. Robinson: A Course in the Theory of Groups.

Let $G$ be a $k$-transitive permutation group of degree $n$ which is neither alternating nor symmetric. Assume $k > 5$. Prove that $(n-k)! \ge 2n$. Deduce that $k \le n-4$.

If $G$ is $k$-transitive, then $n(n-1)\cdots (n-k+1)$ divides $|G|$. Now looking at the first inequality it seems to be related to the order $\frac{(n-k)!}{2} = |A_{n-k}|$ of the alternating group on $n-k$ symbols. Denote by $(G_{\alpha})_G = \bigcap_{g\in G} G_{\alpha}^g$ the normal core of $G_{\alpha}$, as $|G : (G_{\alpha})_G| \ge |G : G_{\alpha}| = n$, if I can embed somehow the factor group $G / (G_{\alpha})_G$ into $A_{n-k}$ the result would follow. This are just some thoughts and I have no idea how to proceed, on what set of size $n-k$ should $(G/G_{\alpha})_G$ act such that it only produces even permutations I do not know. I have no other idea, so I am stuck. Does anyone know how to solve it?
 A: I post here my answer, which is basically Derek's one, so that this question has an explicit, complete answer.
Obviously we can assume that $G$ is a subgroup of $S_n$. Since $k>6$ and $G$ is neither alternating nor symmetric, a Theorem of Jordan (see here) tells us that $G$ is not sharply $k$-transitive. In terms of order it means that $|G|=n(n-1)\dots(n-k+1)m$ with $m>1$. Assume for a contradiction that $(n-k)!<2n$. Say $l=|S_n:G|$. Then $l=(n-k)!/m<2n/m\leq n$. Since $l>2$ by hypothesis, we have that the normal core of $G$ in $S_n$ is trivial and hence $S_n$ is isomorphic with a subgroup of $S_l$, which is impossible. Thus $(n-k)!\geq 2n$.
As for the second part, one has just to notice that $G$ is $n$, $(n-1)$ or $(n-2)$-transitive if and only if $G$ is either $A_n$ or $S_n$. Since this is not the case, we can assume by a contradiction that $k=n-3$. However, from what proved above we get that $2n\leq (n-n+3)!=6$. Then $n\leq 3$, but this is inconsistent with our assumptions. Hence $k\leq n-4$.
