Chain of finite fields Suppose $$f_i=x^p-x-\left [\prod_{j=1}^{i-1} \alpha_j\right]^{p-1} \in \mathbb{F}_p(\alpha_1,\ldots,\alpha_{i-1})[x]$$ is irreducible over $\mathbb{F}_p(\alpha_1,\ldots,\alpha_{i-1}) $, where $\alpha_j$ is a root of $f_j$ with $j<i$. And suppose $\alpha_a$ is a root of $f_a$. How do I prove that $$\mathbb{F}_p(\alpha_1,\ldots,\alpha_a)=\mathbb{F}_p(\alpha_a)?$$ 
It's easy to see that $$\left [\prod_{j=1}^{a-1} \alpha_j\right]^{p-1} \in \mathbb{F}_p(\alpha_a),$$ how do we conclude that $\alpha_j \in \mathbb{F}_p(\alpha_a)$ ?
 A: I think I sort of see how this would go. I'm not completely sure about how the induction continues, but let me show the inductive step $1\to 2$. And list the assumptions I need when we encounter them.
It is well-known that $f_1(x)=x^p-x-1$ is irreducible over the prime field.
More generally, we know that the Artin-Schreier polynomial $x^p-x-a\in K[x]$ is irreducible over the finite field $K$ if and only if 
$$
tr_K(a)=a+a^p+\cdots +a^{p^{m-1}}\neq0. 
$$
Here $m=[K:\Bbb{F}_p]$.
Anyway, $\Bbb{F}_p(\alpha_1)=\Bbb{F}_{p^p}$, and we need to assume that
$tr_{F_{p^p}}(\alpha_1^{p-1})\neq0$ to conclude that 
$$
f_2(x)=x^p-x-\alpha_1^{p-1}
$$
is irreducible over $\Bbb{F}_{p^p}$. Let $\alpha_2$ be a zero of $f_2(x)$.
I claim that $\Bbb{F}(\alpha_2)$ is the extension field of cardinality $p^{p^2}$ (and hence contains $\alpha_1$). To see this it suffices to show that $\alpha_2$ has exactly $p^2$ conjugates under iterated Frobenius automorphism $F$. Let's roll.
Because $f_2(\alpha_2)=0$ we get that
$$
F(\alpha_2)=\alpha_2^p=\alpha_2+\alpha_1^{p-1}.
$$
Applying $F$ to both sides gives
$$
F^2(\alpha_2)=\alpha_2^p+\alpha_1^{p(p-1)}=\alpha_2+\alpha_1^{p-1}+\alpha_1^{p(p-1)}.
$$
Continuing in the same way we get that
$$
F^p(\alpha_2)=\alpha_2+\sum_{j=0}^{p-1}\alpha_1^{p^j(p-1)}=\alpha_2+tr_{\Bbb{F}_{p^p}}(\alpha_1^{p-1}).
$$
The trace is in the prime field, so it is a fixed point of $F$. Repeating the dose leads to
$$
F^{2p}(\alpha_2)=\alpha_2+2tr_{\Bbb{F}_{p^p}}(\alpha_1^{p-1})
$$
and eventually to
$$
F^{p^2}(\alpha_2)=\alpha_2+p\, tr_{\Bbb{F}_{p^p}}(\alpha_1^{p-1})=\alpha_2.
$$
So the number of iterations of $F$ that we need to return to $\alpha_2$ is
a factor of $p^2$. However, our assumption about the non-vanishing
of $tr_{\Bbb{F}_{p^p}}(\alpha_1^{p-1})$ shows that $p$ rounds is not enough.
Therefore $[\Bbb{F}_p(\alpha_2):\Bbb{F}_p]=p^2$ as claimed.
I am optimistic about this approach in the sense that iterating Frobenius to a known zero of an Artin-Schreier polynomial is straightforward. However, it's not clear to me (at least not yet) how the presence of several earlier zeros affects the game. We need to assume that the constant term has non-vanishing trace. Implying that it, too, must generate the extension of degree that is the appropriate power of $p$. My other misgivings are about whether that trace condition is actually satisfied. After all, $\alpha_1$ is determined up to conjugacy, so using a different $\alpha_1$ won't change the trace. Also, this somewhat reminds me about a construction of an iterated tower of extensions of $\Bbb{F}_2$, where there are open problems (the claims have been satisfied only as far as the Fermat numbers are prime). Granted that is a more difficult problem (about primitivity of a product of a sequence of zeros like here). S.D. Cohen from Bristol would know more about that.

Anyway, I propose that you try to prove that $\Bbb{F}_p(\alpha_n)$ is a degree $p^n$ extension of the prime field. Your claims follow from this.
