I'm reading Proposition 10.9 in Section The Real Numbers from textbook Analysis I by Amann/Escher.
and its proof:
In the case $s^n < a$, the authors define $b$ by $b :=\sum_{k=0}^{n-1} \left (\begin{array}{l}{n} \\ {k}\end{array} \right) s^{k}$. On the other hand, they define $b$ slightly different by $b :=\sum^{*}\left(\begin{array}{c}{n} \\ {2 j-1}\end{array} \right) s^{2 j-1}$ in case $s^n > a$. In my below attempt, I found that it is possible to define $b :=\sum_{k=0}^{n-1} \left (\begin{array}{l}{n} \\ {k}\end{array} \right) s^{k}$ in the latter.
My attempt:
Suppose $s^n > a$, we define $b :=\sum_{k=0}^{n-1} \left (\begin{array}{l}{n} \\ {k}\end{array} \right) s^{k} > 0$. Then there is $\varepsilon \in \mathbb{R}$ such that $0<\varepsilon<\left(s^{n}-a\right) / b$ and $\varepsilon \le \min \{1,s\}$. It follows that
$$\begin{aligned}(s-\varepsilon)^{n} &=s^{n}+\sum_{k=0}^{n-1}(-1)^{n-k}\left(\begin{array}{c}{n} \\ {k}\end{array}\right) s^{k} \varepsilon^{n-k} \\ & \geq s^{n}-\sum_{k=0}^{n-1}\left(\begin{array}{c}{n} \\ {k}\end{array}\right) s^{k} \varepsilon^{n-k} \\ &\geq s^{n}-\varepsilon\sum_{k=0}^{n-1}\left(\begin{array}{c}{n} \\ {k}\end{array}\right) s^{k} \\ &= s^n - \varepsilon b\\ &> a \end{aligned}$$
My question: Why do the authors choose a more complicated approach? Is there anything I miss in their proof?