I came across the following proof of Eisenstein's criterion from the book Galois Theory by Joseph Rotman.
Let $f(x) = a_o +a_1x + ... +a_nx^n \in Z [x]$. If there is a prime p dividing $a_j$ for all i < n, but with p not dividing $a_n$ and $p^2$ not dividing $a_o$, then $f(x)$ is irreducible in $\mathbb Q[x]$.
Let $f(x) = g(x)h(x) = (bo + b_1x + ... + b_mx^m)(c_o + c_1x + ... + c_kx^k)$ we may assume that both g and h lie in Z[x]. By hypothesis, $p|a_0 = b_oc_o$ so that $p | b_0$ or $p | c_0$, by Euclid's lemma in Z; since $p^2$ does not divide $a_o$, only one of them is divisible by $p$, say, $p | c_0$ but p does not divide $b_o$. The leading coefficient $a_n = b_mc_k$ is not divisible by p, so that p does not divide $c_k$ (or $b_m$). Let $c_r$ be the first coefficient not divisible by p (so p does divide $c_o, ... , c_{r-1}$). If r < n, then $p|a_n$ and $b_oc_r = a_r - (b_1c_{r-1} + ... + b_rc_o)$ is divisible by p; hence $p | b_oc_n$ contradicting Euclid's lemma (because p divides neither factor). It follows that r = n, hence k = 0, and $h(x)$ is constant. Therefore, f(x) is irreducible.
Shouldn't we conclude that $g(x)$ is constant from $r=n$, because $h(x)$ would have degree $n$ and therefore $g(x)$ would have degree 0, otherwise the degree of $g(x)h(x)>n$. How do we conclude $k=0$?
It’s a typo. It should be either “$m=0$” (instead of $k=0$) or “$k=n$” (instead of $k=0$).
This follows because we know $c_k$ is not divisible by $p$, $k\leq n$, and $r$ is the smallest index such that $c_r$ is not divisible by $p$. Therefore, $r\leq k\leq n$. Since you just established that $r=n$, it follows that $n=r\leq k\leq n$, so $k=n$ and since $m+k=n$, then $m=0$.