P-adic Norm not Complete on Rational Numbers/Proof 2

Proof
Hensel's Lemma is used to prove the existence of a Cauchy sequence that does not converge.

Lemma 1
Let $x_1 \in \Z_{>0}: p \nmid x_1, x_1 \ge \dfrac {p + 1} 2$

Let $k$ be a positive integer such that $k \ge 2, p \nmid k$.

Let $a = x_1^k + p$.

Lemma 2
Let $\map f X \in \Z \sqbrk X$ be the polynomial:
 * $X^k - a$

Lemma 3
Let $\map {f'} X \in \Z \sqbrk X$ be the formal derivative of $\map f X$.

Lemma 4
From Hensel's Lemma there exists a sequence of integers $\sequence {x_n}$ such that:
 * $(1) \quad \forall n: \map f {x_n} \equiv 0 \pmod {p^n}$
 * $(2) \quad \forall n: x_{n + 1} \equiv x_n \pmod {p^n}$

Lemma 5
From Sequence of Consecutive Integers Modulo Power of p is Cauchy in P-adic Norm then:
 * $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$

$\sequence {x_n}$ is a sequence such that for some $c \in \Q$:
 * $\ds \lim_{n \mathop \to \infty} x_n = c$

in $\struct {\Q, \norm {\,\cdot\,}_p}$

From Product Rule for Sequences in Normed Division Ring then:
 * $\ds \lim_{n \mathop \to \infty} x_n^k = c^k$

Hence:
 * $c^k = a \in \Z$.

From Nth Root of Integer is Integer or Irrational then:
 * $c \in \Z$

This contradicts Lemma 2.

So the sequence $\sequence {x_n}$ does not converge in $\struct {\Q, \norm{\,\cdot\,}_p}$.

The result follows.