Multiplicites of roots of $x^{p^k}-x$ ($p$ is prime) in $L[x]$ with $L$ as an extension of $Z_p$

Question

The following is from A. Białynicki-Birula, "Algebra" (the translation is mine).

(Chapter VI, \$6).

Example 1. Let $K$ be a field and let $b \in K, b \neq 0$. Let us consider the polynomial $x^n - b$. We will prove that if $\chi(K) = 0$, then every root of this polynomial has multiplicity equal to 1, and if $\chi(K) = p \neq 0$, then each root of this polynomial has multiplicity equal to $p^m$, where $p^m$ is the greatest power of $p$ such that $p^m \mid n$.

Remark: In the below text the author indeed wrote "see chapter IV, $6, example 1". It is clearly a mistake as there is no such example and that chapter is "Dimension" (of linear space). For this reason I suppose that he wanted to refer to what I quoted above.

(Chapter X. Algebraic elements; \$4. The field of polynomial factorization)

Theorem 4.1. For every field $K$ and for every polynomial $f \in K[x]$ of degree greater than 0 there exists such an extension $L$ of field $K$ that the polynomial $f$ has factorization into linear factors in the ring $L[x]$ .

Example 1. Let $p$ be a prime number and $k$ a natural number. Theorem 4.1 implies that there exists an extension $L$ of the field $Z_p$ such that the polynomial $x^{p^k} - x$ has factorization into linear factors in the ring $L[x]$. Thus in the ring $L[x]$ we have $$x^{p^k}-x = (x-e_1)(x-e_2)...(x-e_q),\ \ \ \ \ \ \ q=p^k.$$ For every two elements $e_i$, $e_j$ such that $i \neq j$ the condition $e_i \neq e_j$ holds and every root of the polynomial $x^{p^k} - x$ in the field $L$ is equal to $e_k$ for some $k = 1, ..., q$ (see chapter IV, $6, example 1) . (...)

How does the statement of the first example allow us to conclude that for each $i, j$ the fact $i \neq j$ implies that $e_i \neq e_j$? I have tried to prove it in several ways but with no success. I tried to establish that $x^n - b$ in some way gathers all the roots so $x^{p^k} - x$ "has exactly one in its disposal" ($p^m \cdot 1 = p^m$), use powers to get $p^m$ same factors in the factorization of $x^{p^k}-x$ and so on but I can't establish a (sufficient) relation between $x^{p^k} - x$ and $x^n - b$ in this case.
All what I obtained is that:

1. every root $a$ of $x^{p^k} - x$ is a root of $x^{p^k} - a$,
2. if $a$ is a root of $x^{p^k} - x$, then this polynomial cannot be factored in $(x-a)^{p^k}$ as polynomials $x^{p^k} - x$ and $x^{p^k}-a$ are different. From some searching on the Internet, I have seen that proofs of the concerned fact generally use derivative, but I read almost whole book and I'm almost sure that he does not use that.

Answer 1

Note: I deleted my initial answer, because it seems I misunderstood the question a bit when I wrote that.

If one is already given the result in Example $1$ from Chapter $6$, then the statement in Example $1$ following Theorem $4.1$ is fairly easy to prove.

The idea is essentially just to factor $f(x)=x^{p^k}-x$ as $x(x^{p^k-1}-1)$. Clearly $0$ is not a root of $x^{p^k}-1$, so $0$ is not a repeated root of $f(x)$. This means exactly one of the $e_i$s, say $e_1$ WLOG, is $0$. This gives us $x^{p^k-1}-1=(x-e_2)(x-e_3)\dots(x-e_q)$, so it is enough to show that $x^{p^k-1}-1$ has no repeated roots. But this is trivial, as the highest power of $p$ dividing $p^k-1$ is obviously $p^0=1$, so by Chapter $6$, Example $1$, all roots of $x^{p^k}-1$ have multiplicity $1$ (remember we are working in an extension of $\mathbb Z_p$), i.e. all of $e_2,e_3,\dots,e_q$ are distinct. It follows that all $e_i$s are distinct, as desired.

Again, apologies for the initial complicated answer and any confusion it caused. The material there was largely irrelevant.