 Solution

Let $N⊴G$ be proper. If $7$ divides $|N|$, then $N$ contains a Sylow $7$-subgroup of $G$, and normality makes it contain all eight of them. Thus $n_7(N)=8$, so the Sylow divisibility theorem givesSince $|N|$ divides 168 and $N$ is proper, this forces $|N|=56$.

Every nontrivial normal subgroup is transitive by normal subgroup orbits in a faithful prime-degree action. HenceBut $N\cap G_x$ is normal in $G_x\cong S_4$, whereas $S_4$ has three, rather than one, Sylow $2$-subgroups. This contradiction proves that no proper normal subgroup has order divisible by seven.

If instead $3$ divides $|N|$, normality makes $N$ contain all 28 Sylow $3$-subgroups of $G$. Hence $n_3(N)=28$, so $28$ divides $|N|/3$ and therefore $84$ divides $|N|$. A proper such subgroup would have order 84, which is divisible by seven and has just been ruled out.

Finally, if $N$ is any nontrivial normal subgroup, the prime-degree orbit argument makes $N$ transitive, so seven divides $|N|$. This is impossible for a proper normal subgroup. Therefore $G$ is simple.