Sequence of Powers of Number less than One/Sufficient Condition

Theorem
Let $x \in \R$.

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

Let $\sequence {x_n}$ be a null sequence.

Then $\size{x} < 1$.

Proof
By Reciprocal of Null Sequence then $\sequence {x_n}$ converges to $0$ $\sequence {\dfrac 1 {x_n} }$ diverges to $\infty$.

By the definition of diverges to $\infty$ then:
 * $\exists N \in \N: \forall n \ge N: \size {\dfrac 1 {x_n} } \gt 1$

In particular, $\size {\dfrac 1 {x_N} } \gt 1$.

By Ordering of Reciprocals then:
 * $\size {x_N} \lt 1$

That is, $\size {x_N} = \size {x^N} = \size {x}^N \lt 1$.

Aiming for a contradiction, suppose $\size {x} \ge 1$.

By Inequality of Product of Unequal Numbers then $\size {x}^N \ge 1^N = 1$, a contradiction.

So $\size {x} \lt 1$ as required.