You can use Sylow's theorem, which is straightforward and your last exercise. But I think it is better to actually prove this result without the use of Sylow's theorem. We will use the same theorem we proved here.

Obviously exists. Now by Cauchy's theorem there is a subgroup . We will show how to prove exists. By the theorem we proved we know that . Since we can form and we see that divides . By Cauchy's theorem there is a subgroup with . If is the natural projection then is a subgroup of with . Just keep on repeating this argument for and so form and this gives us a chain with and .