Integer to Rational Power is Irrational iff not Integer or Reciprocal

Theorem
Let $m$ be a rational number.

Let $n$ be a positive integer.

Then $n^m$ is an irrational number $n^{\left\lvert{m}\right\rvert}$ is not an integer.

Necessary Condition
Let $n^{\left\lvert{m}\right\rvert} \notin \Z$.

$n^m \in \Q$.

We have that:
 * $\left\lvert{m}\right\rvert = 0 \implies n^{\left\lvert{m}\right\rvert} \in \Z$

It follows that:
 * $\left\lvert{m}\right\rvert > 0$

Then:
 * $\left\lvert{m}\right\rvert>0$
 * $m \in \Q$
 * $n^{\left\lvert{m}\right\rvert} > 0$
 * $n^m \in \Q$

imply that
 * $n^{\left\lvert{m}\right\rvert} = n^{p/q} = \dfrac r s$

for some $\left({p, q, r, s}\right) \in \Z_{>0}^4$ where $r$ and $s$ have no common prime factors.

Raising both sides of $n^{p/q} = \dfrac r s$ to the power of $q$ yields:
 * $n^p = \dfrac {r^q} {s^q}$

We have that:
 * $\dfrac {r^q} {s^q} = n^p \in \Z$

Hence $r^q$ is divisible by $s^q$.

By the Fundamental Theorem of Arithmetic, $r^q$ and $s^q$ have the same prime factors as $r$ and $s$ respectively.

But we have that:
 * $s^q \mathrel \backslash r^q$

and:
 * $s \ne 1$

imply that $r$ and $s$ have a common prime factor.

But this would contradict the fact that $r$ and $s$ have no common prime factors.

Therefore $s = 1$.

Then:
 * $n^{\left\lvert{m}\right\rvert} = \dfrac r s$

and:
 * $s = 1$

imply that:
 * $r = n^{\left\lvert{m}\right\rvert} \notin \Z$

From this contradiction it follows that:


 * $n^m \notin \Q$

We have that:
 * $n^m=n^{u/v}$ for some $\left({u, v}\right)\in\Z_{\not=0}$.

Then from the definition of a rational power and the existence of a real $v$th root of $n^u$ it follows that:
 * $n^m=n^{u/v}=(n^u)^{1/v}\in\R$

and so $n^m \notin \Q$ and $n^m \in \R$ imply:
 * $n^m \in \R \setminus \Q$

That is, $n^m$ is an irrational number.

Sufficient Condition
Now let $n^m$ be an irrational number.

That is:
 * $n^m \in \R \setminus \Q$

Then because $n^{-m}$ is the reciprocal of $n^m$:
 * $n^{-m} \in \R \setminus \Q$

So:
 * $n^{\left\lvert{m}\right\rvert} \in \R \setminus \Q$

and so since $\Z \subseteq \Q$:
 * $n^{\left\lvert{m}\right\rvert} \notin \Z$