Sequence of Powers of Number less than One/Necessary Condition/Proof 3

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

By the Archimedean Principle, there exists a natural number $M$ such that:
 * $M > \dfrac 1 {\left({1 - \left\vert{x}\right\vert}\right) \epsilon}$

By the Well-Ordering Principle, there exists a smallest natural number $m$ such that:
 * $\exists N \in \N: m > M \left\vert{x}\right\vert^N$

Note that:
 * $m - 1 \le M \left\vert{x}\right\vert^{N+1}$

By elementary algebra, it follows that:
 * $1 - \dfrac 1 m = \dfrac {m - 1} m \le \left\vert{x}\right\vert < 1 - \dfrac 1 {M \epsilon}$

Hence, $m < M \epsilon$.

Therefore, $\left\vert{x}\right\vert^N < \epsilon$.

By Absolute Value Function is Completely Multiplicative or Modulus of Product, it follows that:
 * $\forall n \in \N: n \ge N \implies \left\vert{x^n}\right\vert = \left\vert{x}\right\vert^n \le \left\vert{x}\right\vert^N < \epsilon$

Hence the result, by the definition of a limit.