Let $L=(\gamma(1+\alpha)/2)^{1/(1-\alpha)}$.
Suppose that $|x_k| < L = (\gamma(1+\alpha)/2)^{1/(1-\alpha)}=((\gamma(1+\alpha)/2)^{\alpha/(1-\alpha)})^{1/\alpha}=((\gamma(1+\alpha)/2)^{-1}(\gamma(1+\alpha)/2)^{1/(1-\alpha)})^{1/\alpha}=((\gamma(1+\alpha)/2)^{-1}L)^{1/\alpha}$
from which we conclude that $|x_k|^{\alpha}(\gamma(1+\alpha)) < 2L$. Also $|x_k| < L=(\gamma(1+\alpha)/2)^{1/(1-\alpha)}$ implies that $\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}>2$. Then $|x_{k+1}|=|x_k|\left|1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right|= |x_k|\left(\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}-1\right)>|x_k|$. If $|x_{k+1}| >L$ then from $|x_k|^{\alpha}(\gamma(1+\alpha)) < 2L$ we manipulate to get $|x_k|\left(\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}-1\right) -L< L-|x_k|$ and finally $|x_k|\left|1-\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}\right| -L< L-|x_k|$. So we conclude that $|x_{k+1}|-L < L - |x_k|$ and then from $|x_{k+1}| >L$ and $|x_k| < L$ we know that $||x_{k+1}|-L| < ||x_k|-L|$.
Supppose that $|x_k| > L = (\gamma(1+\alpha)/2)^{1/(1-\alpha)}=((\gamma(1+\alpha)/2)^{-1}L)^{1/\alpha}$ then $|x_k|^{\alpha}(\gamma(1+\alpha)) > 2L$. Also $0<\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<2$ and $|x_{k+1}|=|x_k|\left|1-\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}\right|<|x_k|$. If $|x_{k+1}| < L$ and $1<\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<2$ then from $|x_k|^{\alpha}(\gamma(1+\alpha)) > 2L$ we manipulate to get $|x_k|\left(\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}-1\right) -L> L-|x_k|$ and finally $|x_k|\left|1-\frac{(\gamma(1+\alpha))}{|x_k|^{1-\alpha}}\right| -L> L-|x_k|$. So we conclude that $|x_{k+1}|-L > L - |x_k|$ and $L-|x_{k+1}| > |x_k|-L$ then from $|x_{k+1}| and $|x_k| > L$ we know that $||x_{k+1}|-L| < ||x_k|-L|$. If $|x_{k+1}| and $0<\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<1$ then $\frac{\gamma(1+\alpha)}{|x_k|^{1-\alpha}}<1$ implies that $|x_k| > 2^{1/(1-\alpha)}L\ge 2L$. And from $|x_{k+1}| we know $||x_{k+1}|-L|.
In summary we have shown that if $|x_k| < L$ then $|x_{k+1}| >|x_k|$. If $|x_k| < L$ and $|x_{k+1}| then $||x_{k+1}|-L|<||x_k|-L|$. If $|x_k| < L$ and $|x_{k+1}| >L$ then $||x_{k+1}|-L|<||x_k|-L|$. If $|x_k| > L$ then $|x_{k+1}| <|x_k|$. If $|x_k| > L$ and $|x_{k+1}| >L$ then $||x_{k+1}|-L|<||x_k|-L|$. If $|x_k| > L$ and $|x_{k+1}| then $||x_{k+1}|-L|<||x_k|-L|$. No matter what the case it is true that $||x_{k+1}|-L|<||x_k|-L|$. Therefore the sequence $|x_k|-L$ converges. Showing that the limit of $|x_k|$ must be $L$ is easy.