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Proof that $a^n=b^n$

I have a question about how to prove that $a^n=b^n$ for all $a,b \in \mathbb{Q}$ and $n \in \mathbb{N}$.
I know that this is true, because all algebraic numbers are algebraic integers and algebraic integers are closed under multiplication. But how would one go about proving this algebraically?
I would appreciate any help.

A:

We have
$$a^n=b^n\iff (a^{ -1})^nb^n=(b^{ -1})^na^n=b^{ -n}(b^{n})^a=b^{ -n}b^n=(b^{ -1})^nb^n\iff b^{ -n}=a^{ -n}$$
where we used the fact that for $\alpha \in \Bbb Q$ the conjugate of $\alpha$ is $\alpha^{ -1}$.
Hence we have that
$$\sqrt[n]{a^n}=\sqrt[n]{b^n}$$
Let $p$ be a prime factor of $a^n-b^n$. Then we have
$$p|\sqrt[n]{a^n}-\sqrt[n]{b^n}$$
But $\sqrt[n]{a^n}-\sqrt[n]{b^n}=a^{\frac{n}{p}}-b^{\frac{n}{p}}$
and $\gcd(a,b) \mid a^{\frac{n}{p}}-b^{\frac{n}{p}}$
So either $p$ divides both of $a^{\frac{n}{p}}-b^{\frac{n}{p}}$ and $\sqrt[n]{a^n}-\sqrt[n]{b^n}$ or $p$ divides \$\sqrt[n]{a^n}-\sq 384a16bd22

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