Sequence of Powers of Number less than One/Necessary Condition/Proof 2

Proof
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

Suppose that:
 * $\exists N \in \N: \left\vert{x}\right\vert^N < \epsilon$

Then the result follows by the definition of a limit, because:
 * $\forall n \in \N: n \ge N \implies \left\vert{x^n}\right\vert = \left\vert{x}\right\vert^n \le \left\vert{x}\right\vert^N < \epsilon$

where either Absolute Value Function is Completely Multiplicative is applied.

It remains to consider the case that:
 * $\forall n \in \N: \left\vert{x}\right\vert^n \ge \epsilon$

By the Archimedean Principle, we can choose a natural number $M$ such that:
 * $\displaystyle M > \frac 1 {\left({1 - \left\vert{x}\right\vert}\right) \epsilon}$

Then, by Sum of Geometric Progression:
 * $\displaystyle M \epsilon \le \sum_{k \mathop = 0}^{M-1} \left\vert{x}\right\vert^k = \frac {1 - \left\vert{x}\right\vert^M} {1 - \left\vert{x}\right\vert} < M \epsilon$

a contradiction.

Hence the result.