Proving that a complex polynomial has no roots on the unit circle 
Let $\sum_{k=0}^{n}a_{k}z^{k}$ be a polynomial of degree $n$ with real
  coefficients satisfying $$a_{0}>a_{1}>....>a_{n-1}>a_{n}>0$$ Prove
  that $p(z)=0$ implies $\left|z\right|>1$.

I have seen similar questions here but none of them proves that solutions can't exist on the unit circle. 
Links for similar questions:
Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.
Let $a_n$ be a decreasing sequence. Prove that the power series $\sum a_n x^n$ has no roots in $A=\{z\in C:|z|<1\}$ 
 A: The proof from Let $a_n$ be a decreasing sequence. Prove that the power series $\sum a_n x^n$ has no roots in $A=\{z\in C:|z|<1\}$ can be modified slightly to show that $|z| \le 1$ is not possible for a zero of $p$:
If $p(z) = 0$ and $|z| \le 1$ then
$$
\begin{align}
 a_0  &= \lvert a_0 - (1-z)p(z) \rvert 
 = \left\lvert  \sum_{k = 1}^{n} (a_{k-1} - a_k)z^k + a_n z^{n+1} \right\rvert \\
 &\underset{(*)}{\le} \sum_{k = 1}^{n} (a_{k-1} - a_k) |z^k| + a_n |z| ^{n+1}  \\
 &\underset{(**)}{\le} \sum_{k = 1}^{n} (a_{k-1} - a_k) + a_n  = a_0 \, .
\end{align}
$$
So equality holds 


*

*in the triangle inequality $(*)$, which means that all complex numbers  $(a_{k-1} - a_k)z^k$ and $a_n z^z$ are zero, or have the same argument (see for example Prove equality in triangle inequality for complex numbers),

*and in $(**)$, which implies $|z|=1$.


The positive factors do not affect the argument, therefore $z, z^2, \ldots, z^n$ must all have the same argument. In particular, $z = z^2/z$ is real and positive, with absolute value $1$, so that $z=1$. But $p(1) \ne 0$, obviously.
