[proofplan] We expand $\beta^p = (\alpha + x^k c)^p$ using the binomial theorem and show that each term of $\beta^p - \alpha^p$ is divisible by $x^{k+1}$. The terms with $i \geq 2$ contribute a factor of $x^{2k} \subseteq x^{k+1}$ (since $k \geq 1$). The single term with $i = 1$ contributes a factor of $p \cdot x^k$, and the hypothesis $\operatorname{char}(R/xR) = p$ guarantees $p \in xR$, yielding the additional factor of $x$ needed. [/proofplan] [step:Write $\beta = \alpha + x^k c$ and expand $\beta^p - \alpha^p$ via the binomial theorem] Since $\alpha \equiv \beta \pmod{x^k}$, there exists $c \in R$ with $\beta = \alpha + x^k c$. The binomial theorem in $R$ gives \begin{align*} \beta^p = (\alpha + x^k c)^p = \sum_{i=0}^{p} \binom{p}{i} \alpha^{p-i} (x^k c)^i, \end{align*} so \begin{align*} \beta^p - \alpha^p = \sum_{i=1}^{p} \binom{p}{i} \alpha^{p-i} (x^k c)^i. \end{align*} [/step] [step:Show each term with $i \geq 2$ is divisible by $x^{k+1}$] For $i \geq 2$, the factor $(x^k c)^i = x^{ki} c^i$ is divisible by $x^{2k}$. Since $k \geq 1$, we have $2k \geq k + 1$, so $x^{2k} \in x^{k+1} R$. Therefore each such term $\binom{p}{i} \alpha^{p-i} x^{ki} c^i \in x^{k+1} R$. [/step] [step:Show the $i = 1$ term is divisible by $x^{k+1}$ using $\operatorname{char}(R/xR) = p$] The $i = 1$ term is $\binom{p}{1} \alpha^{p-1} x^k c = p \cdot \alpha^{p-1} x^k c$. Since $R/xR$ has characteristic $p$, the image of $p \cdot 1_R$ in $R/xR$ is zero, meaning $p \in xR$. Write $p = x d$ for some $d \in R$. Then \begin{align*} p \cdot \alpha^{p-1} x^k c = (xd) \cdot \alpha^{p-1} x^k c = d \alpha^{p-1} c \cdot x^{k+1} \in x^{k+1} R. \end{align*} [guided] This is where the characteristic hypothesis is consumed. The condition $\operatorname{char}(R/xR) = p$ means that $p \cdot 1_{R/xR} = 0$ in the quotient ring $R/xR$. Pulling back to $R$, this says $p \in xR$: the prime $p$ is divisible by $x$ in $R$. Why do we need this? The binomial coefficient $\binom{p}{1} = p$ is not automatically divisible by $x^{k+1}$; it only gives us one factor of $x$ (from $p \in xR$). But combined with the factor $x^k$ already present in $(x^k c)^1$, we obtain $p \cdot x^k \in x \cdot x^k R = x^{k+1} R$. This is precisely the right power. What if $R/xR$ did not have characteristic $p$? Then $p$ might be a unit modulo $x$, and the $i = 1$ term would contribute a factor of only $x^k$, not $x^{k+1}$. The conclusion $\alpha^p \equiv \beta^p \pmod{x^{k+1}}$ would fail. [/guided] [/step] [step:Combine the estimates to conclude $\alpha^p \equiv \beta^p \pmod{x^{k+1}}$] From the previous two steps, every term in the sum $\beta^p - \alpha^p = \sum_{i=1}^{p} \binom{p}{i} \alpha^{p-i} (x^k c)^i$ belongs to $x^{k+1} R$. Therefore their sum lies in $x^{k+1} R$, giving \begin{align*} \beta^p \equiv \alpha^p \pmod{x^{k+1}}. \end{align*} [/step]