From my understanding, you used the theorem that S_n is generated by n-1 transpositions (1,2), (2,3), ..., (n-1, n). You showed that (1,2) and (1,2, ... , n) indeed generate those n-1 transpostions, which shows that . We see that G (a set generated by (1,2) and (1,2,...,n) ) is just the set of permutations of {1,2, ..., n}. By the definition of the symmetric group of degree n, it forces . Thus G=S_n, a symmetric group of degree n.

EDIT (some additional remarks): G must contain (1,2) and (1,2, ... , n). Since G is the group by hypothesis, G must contain a set generated by (1,2) and (1,2, ... ,n). Since our choice of G should be minimal, if we show that a set generated by (1,2) and (1,2, ... ,n) is the group, the proof is completed.