A question regarding the proof of $\gcd(a^m-1, a^n-1) = a^{\gcd(m,n)}-1$ I have a problem trying to understand the proof:

Theorem $\boldsymbol{1.1.5.}$ For natural numbers $a,m,n$, $\gcd\left(a^m-1,a^n-1\right)=a^{\gcd(m,n)}-1$

Outline. Note that by the Euclidean Algorithm, we have
$$
\begin{align}
\gcd\left(a^m-1,a^n-1\right)
&=\gcd\left(a^m-1-a^{m-n}\left(a^n-1\right),a^n-1\right)\\
&=\gcd\left(a^{m-n}-1,a^n-1\right)
\end{align}
$$
original image
We can continue to reduce the exponents using the Euclidean Algorithm, until we ultimately have $\gcd\left(a^m-1,a^n-1\right)=a^{\gcd(m,n)}-1$. $\square$
I see that the next iteration (if possible) is $\gcd(a^{m-2n}, a^n-1)$. However, why is the conclusion is true by euclidean algorithm?

Suppose :
$gcm(m, n)= k$
$m=m_1 k$
$n=n_1 k$
Then we have:
$a^m-1=(a^K)^{m_1}-1=(a^k-1)[(a^k)^{m_1-1}+(a^k)^{m_1-2}+ . . . $
$a^n-1=(a^K)^{n_1}-1=(a^k-1)[(a^k)^{n_1-1}+(a^k)^{n_1-2}+ . . . $
That means:
$gcd(a^m-1, a^n-1)=a^k-1=a^{gcm(m, n)}-1$
Notice that the factor $(a^k-1)$ exists for both cases where m and n are both odd or even.
