I have a question about the following paragraph from Dummit and Foote on separable polynomials:
We now investigate further the structure of inseparable irreducible polynomials over fields of characteristic $p$. We have seen above that if $p(x)$ is an irreducible polynomial which is not separable, then its derivative $D_x p(x)$ is identically $0$, so that $p(x) = p_1(x^p)$ for some polynomial $p_1(x)$. The polynomial $p_1(x)$ may or may not itself be separable. If not, then it too is a polynomial in $x^{p}$: $p_1(x) = p_2(x^p)$, so that $p(x)$ is a polynomial in $x^{p^2}$: $p(x) = p_2(x^p)$. Continuing in this fashion we see that there is a uniquely defined power $p^k$ of $p$ such that $p(x) = p_k(x^{p^k})$ where $p_k(x)$ has nonzero derivative. It is clear that $p_k(x)$ is irreducible since any factorization of $p_k(x)$ would, after replacing $x$ by $x^{p^k}$, immediately imply a factorization of the irreducible $p(x)$. It follows that $p_k(x)$ is separable. We summarize this as:
Proposition 38. Let $p(x)$ be an irreducible polynomial over a field $F$ of characteristic $p$. Then there is a unique integer $k > 0$ and a unique irreducible separable polynomial $p_{sep}(x) \in F[x]$ such that $p(x) = p_{sep}(x^{p^k}).$
I don't understand why this process stops with a separable polynomial (and why does it stop, can't we get stuck with the same polynomial in some sort of an infinite loop ?)