I've to prove that n is generated by the two permutations (1 2), (1 2 3 ... n).
I've already proved that the symmetric group can generated by the following transpositions
(1 2) (2 3) (3 4), ..., (n − 1 n)
in a previous exercise.
So I think I've to prove that those two elements are able to generate all elements of the form (m,m+1) with $m = 1, \ldots n-1$
Note that $$(12)(123 \dots n)=(13)$$ $$(13)(123 \dots n)=(14)$$ $$(14)(123 \dots n)=(15)$$ $$\vdots$$ $$(1 \text{ n -1})(123 \dots n)=(1 \text{ n})$$
Next we do, $$(12)(13)(12)^{-1}=(23)$$ $$(13)(14)(13)^{-1}=(34)$$ $$\vdots$$ $$(1\text{ n-1} )(1\text{ n} ){(1 \text{ n-1} )}^{-1}=(\text{n-1} \text{ n})$$