Let $f(x) \in R[x]$ be a polynomial satisfying Eisenstein's Criterion with prime $p$

Suppose that $f(x)=g(x)h(x)$ with $g(x), h(x) \in F[x]$ where $F$ is the field of fractions of $R$ Gauss' Lemma II states that there exist $g'(x), h'(x) \in R[x]$ such that $f(x)=g'(x)h'(x)$ i.e. any factorization can be converted to a factorization in $R[x]$

Let $f(x) = \sum_{i=0}^n a_i x^i$ $g'(x)= \sum_{j=0}^\ell b_j x^j$ $h'(x)= \sum_{k=0}^m c_k x^k$ be the expansions of $f(x), g'(x)$ and $h'(x)$ respectively.

Let $\varphi : R[x] \rightarrow R/pR[x]$ be the natural homomorphism from $R[x]$ to $R/pR[x]$ Note that since $p \mid a_i$ for $i<n$ and $p \nmid a_n$ we have $\varphi(a_i)=0$ for $i < n$ and $\varphi(a_i)=\alpha \neq 0$ $$ \varphi(f(x)) = \varphi \left(\sum_{i=0}^n a_i x^i \right) = \sum_{i=0}^n \varphi(a_i) x^i = \varphi(a_n) x^n = \alpha x^n $$

Therefore we have $\alpha x^n = \varphi(f(x))=\varphi(g'(x)h'(x))=\varphi(g'(x))\varphi(h'(x))$ so we must have $\varphi(g'(x))=\beta x^{\ell'}$ and $\varphi(h'(x)=\gamma x^{m'}$ for some $\beta, \gamma \in R/pR$ and some integers $\ell', m'$

Clearly $\ell' \leq \deg(g'(x))=\ell$ and $m' \leq \deg(h'(x))=m$ and therefore since $\ell' m' = n = \ell m$ we must have $\ell'=\ell$ and $m'=m$ Thus $\varphi(g'(x))=\beta x^\ell$ and $\varphi(h'(x))=\gamma x^m$

If $\ell>0$ then $\varphi(b_i)=0$ for $i<\ell$ In particular, $\varphi(b_0)=0$ hence $p \mid b_0$ Similarly if $m>0$ then $p \mid c_0$

Since $f(x)=g'(x)h'(x)$ by equating coefficients we see that $a_0 = b_0 c_0$

If $\ell>0$ and $m>0$ then $p \mid b_0$ and $p \mid c_0$ which implies that $p^2 \mid a_0$ But this contradicts our assumptions on $f(x)$ and therefore we must have $\ell=0$ or $m=0$ that is, we must have a trivial factorization. Therefore $f(x)$ is irreducible.