Given a rational number $a/b$ expressed in simplest terms (so $GCD(a,b)=1$), I want to raise it to an integer power $n$.
I think the result will always automatically be in simplest terms, but it's a long time since I was doing maths regularly, so I'm including my reasoning in an answer - please check it.
This question is related, and the answers seem particularly close, but I don't think it's a duplicate.
On
As I say in the question, I think the answer is yes - the result is automatically in least terms.
If $a/b$ is in least terms, $a$ and $b$ don't share any prime factors. If they shared a prime factor $p$, you could reduce the fraction to $a'/b'$ where $a'=a/p$ and $b'=b/p$.
When you calculate $x'=x^n$ for any positive integer $n$, $x'$ has the exact same prime factors as $x$ - each prime factor just repeats $n$ times as often in $x'$.
So if $(a/b)^n = a^n/b^n = A/B$, then $A$ and $B$ share the same prime factors as $a$ and $b$ - just $n$ times as often. And since $a$ and $b$ don't share any prime factors, neither do $A$ and $B$.
To add another proof to yours (which is perfectly correct), but avoiding relying on unique factorization, I would do it as follows:
First, prove that $\gcd(ab,c)$ divides $\gcd(a,c)\cdot\gcd(b,c)$. This can be done noting that, by the Euclidean algorithm, we can find integers $n_1,n_2,m_1,m_2$ such that: $$\gcd(a,c)=n_1a+n_2c$$ $$\gcd(b,c)=m_1b+m_2c.$$ Their product is thus $n_1m_1 ab + n_1m_2 ac + n_2m_1 bc + m_2n_2 c^2$. Each term of this sum is divisible by $ab$ or $c$, thus is divisible by $\gcd(ab,c)$.
Using this, is is clear that $\gcd(a^n,b^n)$ divides $\gcd(a,b)^{n^2}$. If the latter is $1$, so must be the former.