Show that $S_n$ is a maximal subgroup of $S_{n+1}$. Question:

Show that $S_n$ is a maximal subgroup of $S_{n+1}$.

Thoughts: This question is answered here: Maximal subgroup of $S_n$, by adjusting the notation a bit. I was wondering though if there are any other elementary ways to show this.
For example, would it help to show that if $\sigma$ is some cycle in $S_{n+1}-S_n$, then would showing that $S_{n+1}\cup gS_n$?  I was thinking of sort of the idea that goes along with something like, if we have a group, $G$, such that $G$ isn't contained in $A_n$, then $GA_n=S_n$, as long as $G\leq S_n$.
If anyone has any thoughts on this, I would appreciate to hear them!  Thank you!
 A: As elementary as I think is possible:
Suppose $S_n<H\le S_{n+1}$ and fix $h\in H-S_n$, so $h(n+1)=k_h\le n$.
For any $g\in S_{n+1}-S_n$, denote $g(n+1)=k_g\le n$.
$(k_g,k_h)\in S_n$ and $(h^{-1}(k_g,k_h)g)(n+1)=n+1$, so $\tau=h^{-1}(k_g,k_h)g\in S_n$.
Hence $g=(k_g,k_h)h\tau\in H$, showing $H=S_{n+1}$.
A: Here's a proof that avoids any computation (like YCor's) and avoids any difficult concepts (like Robert Chamberlain). Let $H$ be a maximal subgroup containing $K=S_n$. We assume that $K$ is the stabilizer of $n+1$, so any element $g$ of $K\setminus H$ moves $n+1$. But $K$ is transitive on $\{1,\dots,n\}$, so $H$ is transitive on $\{1,\dots,n+1\}$. The orbit-stabilizer theorem now gives $|H|=n\cdot |K|=(n+1)!$.
If you don't even want to use the orbit-stabilizer theorem, you can note that $K^h\leq H$ for all $h\in H$, and since $H$ is transitive, this means all point stabilizers lie in $H$. Thus all transpositions, which stabilize a point for $n\geq 2$, lie in $H$.
A: Say $n\ge 2$. The action of $S_n$ on $\{1,\dots,n\}$ can be identified to that on $S_n/S_{n-1}$ (viewing $S_{n-1}$ as stabilizer of $n$). This action is 2-transitive, hence primitive. And in general, for a group $G$ and subgroup $H$, the action of $G$ on $G/H$ is primitive iff $H$ is a maximal proper subgroup of $G$.
Of course this is not the most straightforward way to see it if one is not familiar with these notions. On the other way, it adapts to other situations and avoids any computation.
