Just need a verification of my proof.
Proof.
Naturally, there arises two cases.
Case 1: $n = a^m$ for some $a \in \mathbb{Z}^+$.
Clearly, $\sqrt[m]{n} = \sqrt[m]{a^m} = a \in \mathbb{Z}^+$.
Case 2: Otherwise.
Suppose that $\sqrt[m]{n} \in \mathbb{Q}$. This would mean that $\exists p,q \in \mathbb{Z}^+$ such that $\sqrt[m]{n} = \frac{p}{q}$ under the conditions that $q \nmid p$ and $\gcd(p,q) = 1$ (otherwise, just simplify).
So, $\sqrt[m]{n} = \frac{p}{q} \implies n = \frac{p^m}{q^m} \implies p^m = nq^m$. Now, notice that by definition of divisibility, $n \mid p^m$. If $n$ is prime, then it follows that $n \mid p$. So, $p = nk_1$ for some positive integer $k_1 \implies (nk_1)^m = nq^m \implies n^{m-1}k_1^m = q^m$. Since $m \geq 2$ (if $m=1$, then $\sqrt[m]{n} = n \in \mathbb{Z}^+$), we have that $m – 1 \geq 1$, so $n \mid q^m$.
Since $n$ is prime, we have that $n \mid q$. But this is a contradiction, since it was assumed that $p,q$ were co-prime but we have just demonstrated that $p,q$ share a common factor of $n$.
Now, if $n$ is not prime, find some prime $p_1$ such that $p_1 \mid n$ (the existence of such a prime is guaranteed by the Fundamental Theorem of Arithmetic). Then, $p_1 \mid p^m$ (since $p_1 \mid n$) $\implies p_1 \mid p$. Thus, by the definition of divisibility, $p = p_1k_2$ for some $k_2 \in \mathbb{Z}^+$. So, making the substitution,
$(p_1k_2)^m = nq^m \implies (p_1k_2)^m = (p_1k_3)q^m$ (since $p_1 \mid n)$ for some $k_3 \in \mathbb{Z}^+$. So, $p_1^{m-1}k_2^m = k_3q^m$.
Using same type of argument as previously, since $m \geq 2, p \mid q^m \implies p \mid q$. Now the contradiction is at hand as $p,q$ share a common factor. QED.
Thanks in advance.
Best Answer
Take for example $m=2$ and $n=18$. Since $n$ is not prime, you want $p_1\mid n$, say $p_1=3$. Following your argument gives $3\mid p$ and $k_3=6$, and your final equation is $$3k_2^2=6q^2\ .$$ This does not prove that $3\mid q$ and so you have no contradiction.
Of course this is easy to fix in this particular case - just take $p_1=2$ instead of $p_1=3$. But you need to work out how to make sure you can do something similar when $n$ is unspecified.
Also consider a case like $m=2$, $n=27$. This is possibly even a bit harder.