Every finite permutation is a product of transpositions, so the transpositions generate $S_n$

For $k\ge2$, pointwise evaluation gives $(a_0\,a_1\,\ldots\,a_{k-1})=(a_0\,a_{k-1})\cdots(a_0\,a_2)(a_0\,a_1)$: the rightmost factor sends $a_0$ to $a_1$, each $a_i$ to $a_{i+1}$, and the leftmost factor sends $a_{k-1}$ back to $a_0$, while all other points are fixed.

Replace each cycle in the decomposition supplied by [L1] with the factorisation in step 1.1 and concatenate the resulting finite lists. This expresses $\sigma$ as a product of transpositions.

If $\sigma$ is the identity, the decomposition and the resulting list are empty. Thus the conclusion includes $n=0$ and $n=1$, and every element of $S_n$ lies in the smallest subgroup containing all transpositions, which by [L2] says that the transpositions generate $S_n$.