I want to know if my proof of the following result is correct:
Let $x$, $y \in Q$, $y > 0$, $x > 1$. Then there is an integer $n$ such that $x^n < y ≤ x^{n+1}$.
Proof:
Suppose I have proved that "let $x$, $y \in Q$, $y > 0$, $x > 1$. Then there is a positive integer $n$ such that $x^n > y$."
By the above result, there is a positive integer $m_0$ such that $x^{m_0} > y$. Consider the set $S = \{n \in \mathbb{N}: y > x^{m_0 - n}\}$. Again, by the above result, there exists a positive integer $m$ such that $1/y < x^m$. Then $x^{-m} < y < x^{m_0}$. Thus, $m_0 + m \in S$ where $m_0 + m \in \mathbb{N}$. By the well-ordering principle, $S$ has a minimal element $m_1$. Then I claim: $x^{m_0 - m_1} < y \leq x^{m_0 - m_1 + 1}$.
Indeed, $m_1 \neq 0$ since $0 \notin S$. Thus, $m_1 - 1 \in \mathbb{N}$, and so if $y > x^{m_0 - m_1 + 1}$, then $m_1 - 1 \in S$, contradicting the minimality of $m_1$. Now we can conclude by letting $n = m_0 - m_1$.