As has been more or less said in comments, I think the important and useful thing to know is that $S_n$ can be generated by two elements.

It is less important which two you choose, but $(1,2)$ and $(1,2,3,\ldots,n)$ has the advantage that it is simply stated uniformly for all $n$.

Perhaps the single most important property of a generating set $X$ of a group $G$, and which could be explained to undergraduates, is that a homomorphism $f:G \to H$ to another group is determined by the images of $f$ on $X$. (This is the same principle as the fact that linear maps are determined by their images on a basis.)

So, if $H$ is a finite group, then there are at most $|H|^{|X|}$ homomorphisms from $G$ to $H$. This is important in particular in both the complexity and practical aspects of algorithmic group theory, which is an active research area.

In fact it embarrassing to have to admit that, there is no known general algorithm for computing ${\rm Aut}(G)$ for finite groups $G$ that has better complexity than the naive method of testing all possible images of the elements in a generating set. (Of course that is not relevant to $G=S_n$, for which ${\rm Aut}(G)$ is known.)

(I should add that, to check whether a given map $X \to H$ really does extend to a homomorphism $G \to H$, you also need a set of defining relations on $X$. There are such sets known for the two "standard generators" of $S_n$ -see here for example - but they are less easily stated.)