1. **Problem statement:** Given that $\gcd(a^n,b^n) = 1$ and $\gcd(a,b^n) = 1$, show that $\gcd(a^{n+1}, b^{n+1}) = 1$. 2. **Recall the properties of gcd:** - $\gcd(x,y)$ divides both $x$ and $y$. - If $\gcd(x,y) = 1$, then $x$ and $y$ are coprime (no common prime factors). 3. **Analyze the given conditions:** - $\gcd(a^n,b^n) = 1$ means $a^n$ and $b^n$ share no common prime factors. - $\gcd(a,b^n) = 1$ means $a$ and $b^n$ share no common prime factors. 4. **From $\gcd(a,b^n) = 1$, deduce $\gcd(a,b) = 1$:** Since $b^n = b \cdot b^{n-1}$, any prime factor of $b$ divides $b^n$. If $a$ and $b^n$ are coprime, then $a$ and $b$ must also be coprime. 5. **Show $\gcd(a^{n+1}, b^{n+1}) = 1$:** Since $a$ and $b$ are coprime, their powers are also coprime. Specifically, for any positive integers $m$ and $k$, if $\gcd(a,b) = 1$, then $\gcd(a^m,b^k) = 1$. Therefore, $\gcd(a^{n+1}, b^{n+1}) = 1$. **Final answer:** $$\boxed{\gcd(a^{n+1}, b^{n+1}) = 1}$$