Coprime powers remain coprime

Prime factorization approach. Since $\divisibility(a,b)=1$, no prime divides both $a$ and $b$. The prime factors of $a^n$ are exactly those of $a$, and similarly for $b^n$. Hence no prime divides both $a^n$ and $b^n$, so $\divisibility(a^n,b^n)=1$.

Bézout approach. From $au+bv=1$, raise to the power $2n-1$:
$$1=(au+bv)^{2n-1}=\sum_{k=0}^{2n-1}\binom{2n-1}{k}(au)^k(bv)^{2n-1-k}.$$
Every term contains either $a^n$ or $b^n$ as a factor (since $k\ge n$ or $2n-1-k\ge n$). So $1$ is a linear combination of $a^n$ and $b^n$, hence they are coprime.