This might not be at the appropriate level for you, but I think this is a cute argument.

Suppose we wanted to transpose $1$ and $n$ in the list $1,\cdots,n$ but we were only allowed to swap adjacent numbers each step, how would we do it? Well, we could use adjacent transpositions to move $1$ all the way to the right, which would shift the numbers $2,\cdots,n$ each one left. Then we could move $n$ to the left with adjacent transpositions until we get the list $n,2,\cdots,n-1,1$.

In other words, we have

$$ (12)(23)\cdots(n-2\,n-1)\cdot(n\,n-1)\cdots(32)(21)=(n1). \tag{$\ast$} $$

(Remember, the rightmost permutations are applied to an input first. Interpret $(u\,v)$ as "swap the numbers that are $u$th and $v$th in the list.") We can use this...

Form a graph as follows: given a generating set $S$ of $H$ consisting of transpositions, let $\{1,\cdots,n\}$ be the vertex set and form an edge $(i,j)$ for every transposition $(ij)\in S$. The hypotheses that $H$ acts transitively and is generated by $S$ is equivalent to saying this graph is connected, so between any two values $a,b\in\{1,\cdots,n\}$ there is some sequence of transpositions

$$ (a_1\,a_2),(a_2\,a_3),\cdots,(a_{k-1}\,a_k)\in S $$

where $a_1=a$ and $a_k=b$. Then we can reuse the $(\ast)$ trick:

$$ (a_1\,a_2)(a_2\,a_3)\cdots(a_{k-2}\,a_{k-1})\cdot(a_k\,a_{k-1})\cdots(a_3\,a_2)(a_2\,a_1)=(a_k\,a_1)=(ab)\in H.$$

That is, $H$ contains every transposition $(ab)$, so it must be $S_n$.

## Best Answer

Any permutation $\sigma \in S_n$ can be written as the product of disjoint cycles. Any cycle $(a_1, \dotsc, a_m) \in S_n$ can be written as the product of transpositions by noting $$ (a_1, a_2, a_3,\dotsc, a_m) = (a_m, a_1)(a_{m-1}, a_1)\dotsc(a_3, a_1)(a_2, a_1). $$ Therefore, the set of all transpositions generate $S_n$. Since $$ (1, j)(1, k)(1, j) = (j, k), $$ the set generated by $(1, 2), \dotsc, (1, n)$ contains all transpositions. Therefore, the set generated by $(1, 2), \dotsc, (1, n)$ is $S_n$.