Prove that the zeros of the polynomial $p(z)=z^n+c_{n-1}z^{n-1}\cdots + c_1z+c_0$ all lie in the open disc centered at $0$ with radius $$R=\sqrt{1+\vert c_0\vert^2+\vert c_1\vert^2+\cdots + \vert c_{n-1}\vert ^2}$$
Attempt: I assume Rouche's Theorem is to be used. Let $g(z)=z^n$. I would like to show $\vert p-g\vert<\vert g\vert$ on the boundary of the disc, where $\vert g \vert =R^n$ (on the boundary). $$\vert p-g\vert \le\vert c_{n-1}z^{n-1}\vert+\cdots+\vert c_1z\vert + \vert c_0\vert\\\le nR^{n-1}\max \vert c_i \vert$$ the latter is the best useful upper bound that I can find, but it is too large since I cannot show that it is less than $R^n$. If it were less than $R^n$, then we'd have $n\max \vert c_i \vert<R$. As a counterexample, let $n=2,c_0=2,c_1=3$, so that $n\max \vert c_i\vert = 2\cdot 3,$ while $R=\sqrt {1+2^2+3^2}=\sqrt{14}.$
How can I find a sharper upper bound for $\vert p-g\vert$ that will allow me to use Rouche's Theorem?
For all $z$ in the circle $|z| = R$, the Cauchy-Schwarz inequality gives
\begin{align}|p(z) - g(z)|& \le \sqrt{|c_0|^2 + |c_1|^2 + \cdots + |c_{n-1}|^2}\cdot \sqrt{1 + R^2 + \cdots + R^{2n-2}}\\ &= \sqrt{R^2 - 1}\cdot \sqrt{\frac{R^{2n} - 1}{R^2 - 1}}\\ &=\sqrt{R^{2n}-1}\\ &< R^n = |g(z)|. \end{align}