Group of order $255$ is cyclic 
Let $G$ a group and its order is $255$.
  Prove that $G$ is cyclic.

I easily demonstrated that the group has only one $17$-Sylow subgroup $P$ that is normal in $G$ and it's cyclic since it is of a prime order. Then $G/P$ is also cyclic since a group of order $15$ is cyclic. Then $G$ can be seen as $G=P(G/P)$ since the orders are coprime and then $G$ is cyclic. Is it correct?

Yes I think it is correct. Because of the orders of all elements must divide to the order of the group and gcd(15,17)=1 you can prove that all the group is generated by only one element (by the same way you prove that a group of order 15 is cyclic)
