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I'm having trouble applying the Euclidean algorithm to $$\gcd(a^n-1,a^m-1)=a^{\gcd(n,m)}-1$$ I've seen others do it such as in this solution but I'm having trouble figuring out what they are doing in each step.

Could someone help explain this to me?

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Assume that $n=qm+r$. Then $ a^n-1 = a^r\cdot(a^m)^q-1 $, and since $$ a^m \equiv 1 \pmod{a^m-1} \tag{1}$$ we have: $$ (a^n-1) \equiv (a^r-1)\pmod{a^m-1}\tag{2} $$ proving: $$ \gcd(a^n-1,a^m-1) = \gcd(a^r-1,a^m-1).\tag{3}$$ Now, repeat. The involved exponents transform like in the computation of $$ \gcd(n,m) = \gcd(r,m) = \ldots \tag{4} $$ hence at last we get: $$ \gcd(a^n-1,a^m-1) = a^{\gcd(n,m)}-1\tag{5} $$ qed.

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