All the zeroes of $p(z)$ lie inside the unit disk Let $p(z) = c_0 + c_1z + c_2z^2 + \dots + c_nz^n$ where $0 \le c_0 \le c_1 \le \dots \le c_n$.  I would like to show that all zeroes of this polynomial lie inside the unit disk by applying Rouche's theorem to the polynomial $(1-z)p(z)$.  I'm not completely sure how to do this. Using the given, information I can deduce that 
$$|(1 - z)p(z)| \le |c_nz^{n+1}|$$
on the unit circle, but this doesn't really match the assumptions of Rouche's theorem. 
Help would be appreciated.
 A: Similar to @DonAntonio's answer but uses Rouché's theorem. We have:
$$
(1 - z)p(z) = c_0 + \sum_{k=1}^n (c_k - c_{k-1}) z^k - c_n z^{n+1}
$$
Let $|z| = r > 1$, we have:
\begin{align}
\left|(1 - z)p(z) - (-c_n z^{n+1})\right| &= \left| c_0 + \sum_{k=1}^n (c_k - c_{k-1}) z^k \right| \\
&< \left(c_0 + \sum_{k=1}^n (c_k - c_{k-1})\right)\left|-z^{n+1}\right| = \left|-c_n z^{n+1}\right|
\end{align}
Thus, $(1 - z)p(z)$ and $-c_n z^{n+1}$ have the same number of zeros inside every circle $|z| = r$ for $r > 1$. But $-c_n z^{n+1}$ has $n+1$ zeros at $0$ and $(1-z)p(z)$ has $n+1$ zeros. It follows that all of the zeros of $(1-z)p(z)$ lie inside the circle $|z| = r$.
By letting $r \to 1$, we conclude that all zeros of $(1 - z)p(z)$ (and hence $p(z)$) lie in the closed unit disk. Notice that zeros can be on the unit circle as demonstrated by the polynomial $p(z) = 1 + z$.
A: Define
$$t(z):=(1-z)p(z)=\sum_{k=1}^n(c_k-c_{k-1})z^k+c_0-c_nz^{n+1}$$
Let $\,w\in\Bbb C\;,\;\;|w|>1\,$ be a root of $\,p(z)\,$ and, thus, also of $\,t(z)\,$:
$$c_n|w^{n+1}|=\left|c_0+\sum_{k=1}^n(c_k-c_{k-1})w^k\right|\le c_0+\sum_{k=1}^n(c_k-c_{k-1})|w|^k<$$
$$<c_0|w|^n+\sum_{k=1}^n(c_k-c_{k-1})|w|^n=c_n|w|^n$$
so we got:
$$c_n|w|^{n+1}<c_n|w|^n\Longrightarrow|w|<1\ldots\text{contradiction}$$
Polish and end the argument now.
