I couldn’t solve this question and I don’t know how to figure this out from beginning. Hint from my school was to think about euclid method on $ α^m-1,α^n-1$,and better to think $α$ as Polynomial. Here is the question
***There are positive integer m,n and $gcd(m, n) = d$ . At this time, show that
$gcd (a^m -1,a^n −1) = a^d −1$ is valid for 2 or more integer a.***
Given $\quad gcd (a^m -1,a^n −1) = a^d −1\quad$ we begin by showing
$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+a^{n-3}+ ...+a^{0})$$ This is true no matter what the power of $a$ so $(a-1)$ is a divisor of any such variable or function. As we can see, however, the cofactor contains $n$ terms with the highest being $a^{n-1}$ so the GCD of the cofactors is $1$. This means that the only divisor of any $(a^x-1)$ is $(a-1)$ or its cofactor for any power $x$. $$\therefore GCD(a^m-1,a^n-1)=(a-1)\text{ where } a,m,n\in\mathbb{N}\text{ i.e. they natural numbers}$$ The rest is merely showing that $a=a^d$ where $d=0$.
