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 $q \in \mathbf Z$.