I found the following theorem in a paper by Yann Bugeaud (page 12) ,

the theorem was not written in detail, to be specific,following two lines on page 13 were not understandable-

I think this theorem may exist in other books, paper, it seems general result (I could be wrong though).
Where can I get the detail version of the proof ? or can anyone prove (providing the detail) above two lines in detail? **or is there an error /typo?** Because a member, Gerhard Paseman, suggested in his answer that-

As of this date, as written, I still think the question is not good for this forum. However, it raises one which I think is a good one for this forum. Here's the good question: Does (16) have a typo?

**PS:**

One of the member tried to prove it but failed, his answer is given as below-

If we divide both sides of equation (15) of page 12 by $\frac{\gamma}{\gamma-\delta} \gamma^{sd}$, we get- $\frac{\alpha(\gamma-\delta)}{\gamma(\alpha-\beta)}{\left(\frac{\gamma^s}{\alpha^r}\right)}^{-d}-1=\frac{\gamma-\delta}{\alpha-\beta}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}-\left(\frac{\delta}{\gamma}\right)^{1+sd}$. Then, we have $\left|\frac{\alpha(\gamma-\delta)}{\gamma(\alpha-\beta)}{\left(\frac{\gamma^s}{\alpha^r}\right)}^{-d}-1\right|=\left|\frac{\gamma-\delta}{\alpha-\beta}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}-\left(\frac{\delta}{\gamma}\right)^{1+sd}\right|$
Since $\frac{\delta}{\gamma}\approx 0$ (i.e. $0<\frac{\delta}{\gamma}<1$ ) from $\gamma\ge |\delta|^{1+\eta}\gg|\delta|$, we have
$\left|\frac{\gamma-\delta}{\alpha-\beta}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}-\left(\frac{\delta}{\gamma}\right)^{1+sd}\right|\approx \left|\frac{\gamma-\delta}{\alpha-\beta}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}\right|$. Note that as the values of $\gamma$ become larger and larger, $\frac{\delta}{\gamma}$ become smaller and smaller and become close to $0$, so, we ignored $\frac{\delta}{\gamma}$ by writing $\frac{\delta}{\gamma}=0.$
Similarly, $\frac{\delta}{\gamma}\approx 0$ and $\frac{\beta}{\alpha}\approx 0$ from $\alpha\ge |\beta|^{1+\eta}\gg |\beta|$, we have, $\frac{\gamma-\delta}{\alpha-\beta}=\frac{1-\frac{\delta}{\gamma}}{1-\frac{\beta}{\alpha}}\cdot \frac{\gamma}{\alpha}\approx\frac{1-0}{1-0}\cdot\frac{\gamma}{\alpha}=\frac{\gamma}{\alpha}$ and so
$\left|\frac{\gamma-\delta}{\alpha-\beta}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}\right|\approx \left|\frac{\gamma}{\alpha}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}\right|$. Since $\alpha^r\approx \gamma^s$ **(how?)**, we have
$\left|\frac{\gamma}{\alpha}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}\right|\approx \left|\frac{\gamma}{\alpha}\cdot \frac{\beta^{1+rd}}{\gamma\cdot \alpha^{rd}}\right|=\frac{\gamma}{\alpha}\cdot \frac{|\beta|^{1+rd}}{\gamma\cdot \alpha^{rd}}=\frac{|\beta|^{1+rd}}{\alpha^{1+rd}}$. **By inspection**, $\alpha\ll \alpha^{1+(1-\eta)}$ and from $\alpha\ge |\beta|^{1+\eta}\gg |\beta| \implies \alpha^{rd}\gg |\beta|^{rd} $ so we get,
$\frac{|\beta|^{1} \beta|^{rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1}|\beta|^{rd}}{\alpha^{1+rd}}$
$\implies \frac{|\beta|^{1} \beta|^{rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1}|\beta|^{(1-\eta)rd}}{\alpha^{1+rd}}$**(how?)**
$ \implies \frac{|\beta|^{1+rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1+(1-\eta)rd}}{\alpha^{1+rd}}=\alpha^{-\eta rd}$.

This answer is taken from another member which is not complete.

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