$a + b$, $b^{p - 1}$ coprime when $a$, $b$ coprime for odd prime $p$

Question

I was reading a proof of the theorem $$\gcd\bigg(\frac{a^p + b^p}{a + b}, a + b\bigg) \in \{1, p\}$$ where $a$, $b$ are coprime integers and $p$ is an odd prime. Using long division, we get $$\gcd(a + b, pb^{p - 1})$$ which equals $\gcd(a + b, p) \in \{1, p\}$ because $a + b$ and $b^{p - 1}$ are coprime.

I don't understand how that's true though. I tried using the Euclidean algorithm but still don't see how $\gcd(a + b, b^{p - 1}) = 1$.

Answer 1

$\gcd(a+b,b^p)|\gcd(a+b,b)^p=1^p=1$