Assum that $a\geq 1$. Let
$S_{a}=\left\{ x\in \mathbb{R} :x^{n}\leq a\right\} $.
Then $S_{a}$ is a subset of $\mathbb{R}$, which is nonempty (since $1\in S_{a}$) and bouded above. Define $b=\sup S_{a}$. Note that since $1\in S_{a}$, we have $b\geq 1>0$.
For any $\delta\in \mathbb{R}$, we have
$\left( b+\delta \right) ^{n}=b^{n}+\left( \begin{matrix} n\\ 1\end{matrix} \right) b^{n-1}\delta+\left( \begin{matrix} n\\ 2\end{matrix} \right) b^{n-2}\delta^{2}+\ldots +\delta^{n}$.
Now, suppose $b^{n} < a$. Let us define
$\epsilon :=a-b^{n}$, $M:=\max \left\{ \left( \begin{matrix} n\\ k\end{matrix} \right) b^{n-k}:k=1,2,\ldots ,n\right\} $ and $\delta:=\min \left\{ 1,\dfrac {\epsilon } {nM}\right\} $.
I want to know why $0 < \delta ^{k}\leq \delta$ in the proof.