The statement to be proven is: "Let $r$ be a rational number and $a$ an integer such that $r^n=a$. Prove that $r$ is an integer."
Proof: Let $a = (\frac{b}{c})^n$ where $b,c \in \mathbb{Z}$. Assume that $gcd(b,c)=1$. Then $a^{\frac{1}{n}} = \frac{b}{c}$. This implies that $b=ca^{\frac{1}{n}}$ which implies that $b \vert ca^{\frac{1}{n}}$ Since $gcd(b,c)=1$ we must have that $b \vert a^{\frac{1}{n}}$. This implies that $a^{\frac{1}{n}} = bs$ for some $s \in \mathbb{Z}$. Since $bs \in \mathbb{Z}$ this implies that $a^{\frac{1}{n}} \in \mathbb{Z}$. Thus $r=a^{\frac{1}{n}} \in \mathbb{Z}$.
If something is wrong I would greatly appreciate a pointer in the right direction!
Edit: I realized from the replies below that my argument was circular. Here is my new attempt should anyone else stumble across this:
Proof: Suppose $r=\frac{b}{c}$. Then $r^n=a$ implies that $(\frac{b}{c})^n = a$ so that $b^n = ac^n$. This implies that $c^n \vert b^n$. Since $c^n$ and $b^n$ have no common prime factors, $c^n \vert b^n$ is impossible unless $c^n=1$, which implies $c=1$ or $c=-1$. This implies that $r=b$ or $r=-b$ so that $r \in \mathbb{Z}$.
I would still appreciate any feedback :)
No, it is not a valid proof. You have $b|ca^\frac{1}{n}$; but you can't deduce from this that $b|a^\frac{1}{n}$ unless you already know that $a^\frac{1}{n}$ is an integer. For instance, $3$ divides $7\cdot\frac{6}{7}$, and $\gcd(3,7)=1$; but it is nonsense to deduce from this that $3$ divides $\frac67$.