How many subgroups of order 17 does $S_{17}$ have ?
My attempt :
An order 17 group is of prime order, hence cyclic and each element in it is a generator and of order 17.
In $S_{17}$ group we can get an order 17 element only through a 17-cycle.
Number of elements of order 17 in $S_{17}$ is $\frac{17!}{17} = 16!$.
Now given that two sylow 17 subgroups have only a trivial intersection. We can conclude that 16 of these elements fall into each sylow 17 subgroup.
Hence the number of sylow 17 subgroups would be $\frac{16!}{16} = 15!$
Your answer looks correct.
It is true that there are $16!$ elements of order $17$, but if this is a homework question a marker might want you to elaborate on why that is.
An alternative (but not necessarily better) proof is as follows:
Consider the set of subgroups of $S_{17}$ of order $17$. As you noted, these subgroups must be cyclic. In particular they must be transitive.
For each subgroup $G$ fix $g\in G$ with $g(1)=2$. $g$ is the only such element of $G$ and generates $G$ so uniquely defines $G$.
Write $g=(1,2,x_3,\ldots,x_{17})$. There are $15!$ choices for the $x_i$ so $15!$ such subgroups.