Are you missing the following bits?
- The polynomial $f(x)\in \mathbb{Z}_p[x]$ of degree $n$ has a root in the field $E$. Because the field extension $E/\mathbb{Z}_p$ is Galois, and $f(x)$ is irreducible, all the roots of $f(x)$ are distinct and in $E$. So $ f(x)=(x-a_1)(x-a_2)\cdots (x-a_n) $ for some elements $a_1,a_2,\ldots,a_n\in E$.
- The elements of $E$ are exactly the zeros of the polynomial $p(x)=x^{p^n}-x$. In other words $ p(x)=\prod_{a\in E}(x-a). $
The troubling claim follows from this. The zeros $a_i,i=1,2,\ldots,n,$ are among the zeros of $p(x)$, so $f(x)\mid p(x)$.
In particular, the polynomial $f(x)$ also has $n$ zeros in $E'$, because $p(x)$ has $p^n$ roots there, and the roots of $f(x)$ are among those.
Edit: Proving my first claim. This depends heavily on the properties of the so called Frobenius homomorhpism $F:E\to E, x\mapsto x^p$. This is a homomorphism, because obviously $F(1)=1$ and $F(xy)=(xy)^p=x^py^p=F(x)F(y)$ for all $x,y\in E$. Less obvious is that $F$ respects addition as well, i.e. $ F(x+y)=(x+y)^p=x^p+y^p=F(x)+F(y) $ for all $x,y\in E$. This follows from the binomial formula together with the observation that the binomial coefficients ${p\choose i}$ are all divisible by $p$, when $1\le i\le p-1$.
From little Fermat it follows that $F(x)=x^p=x$ for all the elements $x$ of the subfield $\mathbb{Z}_p$. We need to also make the observation that $x^p=x$ only when $x\in\mathbb{Z}_p$. This is because the polynomial equation $x^p-x=0$ can have at most $p$ solutions in the field $E$, and we already found $p$ solutions.
So we assume that $f(x)=x^n+f_{n-1}x^{n-1}+f_{n-2}x^{n-2}+\cdots+f_1x+f_0\in \mathbb{Z}_p[x]$ is irreducible, and has a root $a_1$ in $E$ (=the coset of $x$ in $\mathbb{Z}_p[x]/\langle f(x)\rangle$). In other words $a_1\notin\mathbb{Z}_p$ and $ a_1^n+f_{n-1}a_1^{n-1}+\cdots f_1a_1+f_0=0. $ Let's apply the mapping $F$ to this equation. Remember that $F(f_i)=f_i$ for all $i$. We get $ a_1^{pn}+f_{n-1}a_1^{p(n-1)}+\cdots f_1a_1^p+f_0=0, $ or, upon inspection, $f(a_1^p)=0$. Because $a_1\notin\mathbb{Z}_p$, $a_1^p\neq a_1$. Therefore we have found another zero $a_2=a_1^p$ of $f(x)$ in $E$.
We can repeat the argument and keep finding roots of $f(x)$: $a_3=a_2^p$, $a_4=a_3^p$ et cetera. Because $f(x)$ can have at most $n$ roots in $E$, this sequence of roots will have to start repeating at some point. Because $F$ is injective (its kernel is trivial), the repetition must start from $a_1$, in other words $a_1=a_1^{p^k}$ for some $k, 2\le k\le n$.
The polynomial $ g(x)=(x-a_1)(x-a_1^p)\cdots (x-a_1^{p^{k-1}}) $ is stable under $F$, so its coefficients are in $\mathbb{Z}_p$. Furthermore, $g(x)\mid f(x)$. But $f(x)$ was irreducible, so we must have $g(x)=f(x)$, and $k=n$. But all the roots of $g(x)$ are distinct and in $E$ by construction. Therefore the same holds for $f(x)=g(x)$.