# Showing the existence of $b$ such that $a = b^p$

I have the following question:

Let $$F$$ be a field of characteristic $$p$$ (where $$p$$ is prime) such that every irreducible polynomial in $$F[x]$$ is separable. Show that for every $$a\in F$$, there is $$b\in F$$ such that $$a=b^p$$

I have shown that in a splitting field $$E$$, for the polynomial $$x^p -a$$, there can be only one root, and therefore, this polynomial is reducible. I'm stuck on how to finish the proof.

Here's a recap of the work I've done. If $$E$$ is the splitting field of $$x^p-a$$ over $$F$$, then char$$E$$ = char$$F$$ = p. If $$b$$ and $$c$$ are both roots of this polynomial in $$E$$, then it must be the case that $$b=c$$, so in the splitting field, $$x^p -a = (x-b)^p = x^p-b^p$$ by the Freshman's Dream". Hence, we know by assumption that $$x^p -a$$ is actually reducible in $$F$$. However, this is where I'm stuck.

Any help would be appreciated, especially in determining a next step. TIA.

Write $$X^p-a$$ has a product $$X^p-a=P_1(X)....P_n(X)$$ where $$P_i(X)$$ is irreducible, suppose that $$deg(P_1)>1$$, there exists an extension $$E$$ of $$F$$ which is the splitting field of $$P_1$$ let $$b$$ be a root of $$P_1$$ in $$E$$, we deduce that $$P_1(b)=0$$ and $$b^p-a=P_1(b)...P_n(b)=0$$, this implies that $$X^p-a=(X-b)^p$$ since $$P_1$$ divides $$X^p-a=(X-b)^p$$ it is not separable contradiction.