That is easy to see. For the second inequality, . The latter can be shown using the binomial theorem, according to which .

In fact, the whole power of the theorem is not needed. One can write as . When performing the multiplication, we have to pick either 1 or x from each factor, and consider and add all such variants. There is only one way to choose 1 from every factor; all other terms will have x in them. Then we can choose x from the first factor and 1's from all the rest, or x from the second factor and 1's from all the rest, etc. In the sum, this will give us . All the rest terms in the sum will have at leat .