Field of characteristic p - perfect iff pth roots of elements are all in the field.

Prove the following theorem: A field of characteristic p is perfect if only if there exists within the field a pth root for every element.

In the book I'm using(van der Waerden's algebra) the author proves that every irreducible polynomial in such a field is a pth power(easy). But I don't understand why it is separable if it's a pth power. What am I doing wrong?

The whole point is the following: You prove that every irreducible polynomial, which is not separable, is a $p$-th power. But that is a contradiction, because the polynomial was irreducible.