number theory - How to prove that any (integer)$^{1/n}$ that isn't an integer, is irrational?

Is my proof beneath perfect and complete?

I wanted to prove that for any nth root of an integer, if it's not an integer, than it's irrational: $$\begin{cases} m,n\in \mathbb{N}\\\sqrt[n]{m}\notin \mathbb{N} \end{cases}\implies \sqrt[n]{m}\notin \mathbb{Q}.$$

I start by assuming that $m^{\frac 1n}$ is rational and non-integer. So there exist co-prime integers $a,b$ so that $$\sqrt[n]{m}=\frac{a}{b}$$ $$\implies m=\frac{a^n}{b^n}\in\mathbb{N}.$$ But since $a$ and $b$ have no common factor, $a^n$ and $b^n$ also have no common factor. So: $$\frac{a^n}{b^n}\notin\mathbb{N},$$ a contradiction.

Answer

Your proof is fine. You can use essentially the same idea to prove the following more general statement:

Theorem. If $ P(X) \in \mathbf Z[X] $ is a monic polynomial, then any rational roots of $ P $ are integers. In other words, $ \mathbf Z $ is integrally closed.

Proof. Assume that $ q = a/b $ is a rational root with $ a, b $ coprime, and let $ P(X) = X^n + c_{n-1} X^{n-1} + \ldots + c_0 $. We have $ P(q) = 0 $, which gives

$$ a^n + c_{n-1} a^{n-1} b + \ldots + c_0 b^n = 0 $$

In other words, $ a^n $ is divisible by $ b $. This is a contradiction unless $ b = \pm 1 $, since then any prime dividing $ b $ also divides $ a $, contradicting coprimality. Hence, $ b = \pm 1 $ and $ q \in \mathbf Z $.