Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Cauchy Sequence Equivalent

Theorem
Let $R$ be a division ring.

Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$. Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
 * $\exists \alpha \in \R_{> 0}: \forall x \in R: \norm x_1 = \norm x_2^\alpha$

Then for all sequences $\sequence {x_n}$ in $R$:
 * $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$

Proof
Let $\sequence {x_n}$ be a Cauchy sequence in $\norm {\, \cdot \,}_1$.

Let $\epsilon > 0$ be given.

Since $\sequence {x_n}$ is a Cauchy sequence then:
 * $\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_1 < \epsilon^\alpha$

Then:
 * $\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon^\alpha$

Hence:
 * $\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 < \epsilon$

So $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$

It follows that for all sequences $\sequence {x_n}$ in $R$:
 * $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$

Let $\sequence {x_n}$ be a Cauchy sequence in $\norm {\, \cdot \,}_2$.

Let $\epsilon > 0$ be given.

Since $\sequence {x_n}$ is a Cauchy sequence then:
 * $\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 \lt \epsilon^{1/\alpha}$

Then:
 * $\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon$

Hence:
 * $\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_1 < \epsilon$

So $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$

It follows that for all sequences $\sequence {x_n}$ in $R$:
 * $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$

The result follows.