Sequence of Powers of Number less than One/Normed Division Ring

Theorem
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring

Let $x \in \R$.

Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.

Then:
 * $\norm x < 1$ $\sequence {x_n}$ is a null sequence.

Proof
Let $0_R$ be the zero of $R$. By the definition of convergence:
 * $\displaystyle \lim_{n \mathop \to \infty} x_n = 0_R \iff \lim_{n \mathop \to \infty} \norm {x_n} = 0$

By norm axiom (N2) (Multiplicity) then for each $n \in \N$:
 * $\norm {x_n} = \norm {x^n} = \norm x^n$.

So:
 * $\displaystyle \lim_{n \mathop \to \infty} \norm {x_n} = 0 \iff \lim_{n \mathop \to \infty} \norm x^n = 0$

Since $\norm x \in \R_{\ge 0}$, by Sequence of Powers of Real Number less than One:
 * $\displaystyle \lim_{n \mathop \to \infty} \norm x^n = 0 \iff \norm x < 1$

The result follows.