Let \(a=r_0\), \(b=r_1\) and \(r_{i+2}\) be the remainder after dividing \(r_i\) by \(r_{i+1}\):
\begin{align} r_0 =& r_1 q_1 + r_2 \\ r_1 =& r_2 q_2 + r_3 \\ &\vdots \\ r_{n-2} =& r_{n-1}q_{n-1} \end{align}
where \(r_{n-1} = (r_0,r_1)\). According to this lemma, if \( r_{i+2}\) is the remainder after dividing \(r_i\) by \(r_{i+1}\), then \(2^{r_{i+2}} - 1\) is the remainder after dividing \(2^{r_i} - 1\) by \(2^{r_{i+1}}-1\). This means:
\begin{align} 2^{r_0}-1 =& (2^{r_1}-1) h_1 + 2^{r_2}-1 \\ 2^{r_1}-1 =& (2^{r_2}-1) h_2 + 2^{r_3}-1 \\ &\vdots \\ 2^{r_{n-1}}-1 =& (2^{r_{n-2}}-1) h_{n-2} \end{align}
This shows that \(2^{r_{n-2}}-1 = (2^{r_0}-1,2^{r_1}-1)\), where \(r_{n-2} = (r_0,r_1)\).