$$\begin{align*}
|a_{n-1}|\cdot |z^{n-1}| + \cdots + |a_1|\cdot| z| + |a_0 |
&= |a_{n-1}|\cdot R^{n-1} + \cdots + |a_1| R + |a_0 | \\
&\leq |a_{n-1}|\cdot R^{n-1}+|a_{n-2}|\cdot R^{n-1} + \cdots + |a_1| \cdot R^{n-1} + |a_0 |\cdot R^{n-1} \qquad\text{(1)}\\
&= R^{n-1} \left( |a_{n-1}| + |a_{n-2}| + \cdots + |a_1| + |a_0| \right) \\
&\leq R^{n-1} \left( |a_n|\cdot R \right) \qquad\text{(2)}\\
&= R^{n} |a_n|
\end{align*}
$$
In inequality (1) I have used the fact that $R^{i} \leq R^{n-1}$ for $0 \leq i \leq n-1$ and inequality (2) follows by choice of $R$.
Then he claims that "$p(w+\varepsilon e^{i\theta})=p_0+\color{red}{p_m\varepsilon^{mi\theta}}+O(\varepsilon^{m+1})$
I assume that was meant to be $\,p_m\color{red}{\varepsilon^m}\varepsilon^{mi\theta}\,$ instead. Then:
$$
\begin{align}
\left|p(w+\varepsilon e^{i\theta})\right|^2 &= p(w+\varepsilon e^{i\theta})\,\overline{p(w+\varepsilon e^{i\theta})} \tag{1}
\\ &= \left(p_0+p_m\varepsilon^m\varepsilon^{mi\theta}+O(\varepsilon^{m+1})\right)\,\left(\overline{p_0+p_m\varepsilon^m\varepsilon^{mi\theta}}+O(\varepsilon^{m+1})\right) \tag{2}
\\ &= \left|p_0+p_m\varepsilon^me^{mi\theta}\right|^2+O(\varepsilon^{m+1})\cdot\left(O\left(\varepsilon^0\right)+O\left(\varepsilon^0\right)+O\left(\varepsilon^{m+1}\right)\right) \tag{3}
\\ &= \left|p_0+p_m\varepsilon^me^{mi\theta}\right|^2+O(\varepsilon^{m+1}) \tag{4}
\end{align}
$$
[ EDIT ] $\;$ Step by step:
$\;\;{(1)}\;$ using that $\,|z|^2 = z \,\overline{z}\,$ for all complex $\,z\,$;
$\;\;{(2)}\;$ using that $\,\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\,$ and $\,\overline {\,O(\varepsilon^k)\,} = O(\varepsilon^k)\,$ for real $\,\varepsilon\,$;
$\;\;{(3)}\;$ using that $\,a + b \cdot \varepsilon^k+\dots = a \color{red}{\cdot \varepsilon^0} + b \cdot \varepsilon^k+\dots=O\left(\varepsilon^0\right)\,$;
$\;\;{(4)}\;$ using that $\,O(n_1)\,\pm\,O(n_2) = O\left(\min (n_1,n_2)\right)\,$ and $\,O(n_1)\,O(n_2) = O(n_1+n_2)\,$.
Best Answer
Suppose $|z| > 2\max_i\left|{\large{\frac{p_i}{p_n}}}\right|$.
In particular, using $i=n$, we get $|z| > 2$, hence $2(|z|-1) > |z|$.
Letting $r=\max_i |p_i|$, we have $|z| > {\Large{\frac{2r}{|p_n|}}}$, hence \begin{align*} &\left|\sum_{i=0}^{n-1}p_iz^i\right|\\[4pt] \le\;&\sum_{i=0}^{n-1}|p_iz^i|\\[4pt] =\;&\sum_{i=0}^{n-1}|p_i||z|^i\\[4pt] \le\;&r\sum_{i=0}^{n-1}|z|^i\\[4pt] =\;&r\left(\frac{|z|^n-1}{|z|-1}\right)\\[4pt] =\;&2r\left(\frac{|z|^n-1}{2(|z|-1)}\right)\\[4pt] < \;&2r\left(\frac{|z|^n}{|z|}\right)\\[4pt] =\;&2r|z|^{n-1}\\[4pt] =\;&|p_n|\cdot \frac{2r}{|p_n|}\cdot |z|^{n-1}\\[4pt] < \;&|p_n|\cdot |z|\cdot |z|^{n-1}\\[4pt] =\;&|p_nz^n|\\[4pt] \end{align*}