Prove that $gcd(\frac{a^n-b^n}{(a-b)}, a-b) = gcd(n\cdot gcd(a,b)^{n-1}, a-b)$ is true for a,b $\in$ $\mathbb{Z}^+$
I can see how it could be true. I know $\frac{a^n-b^n}{(a-b)} = a^{n-1} + ba^{n-2} + ... + b^{n-1}$
Then $gcd(a,b)^{n-1}$ is going to be of the same order as $a^{n-1} + ba^{n-2} + ... + b^{n-1}$ since this polynomial is homogeneous. I do not know why there is an n being multiplied out in front of $gcd(a,b)^{n-1}$. Using the Euclidean Algorithm would be kind of difficult with n-cases where n is unknown, even trying to use induction.
