Proving that $\sum\limits_{n = 0}^{2013} a_n z^n \neq 0$ if $a_0 > a_1 > \dots > a_{2013} > 0$ and $|z| \leq 1$

You can look at (with $N$ = 2013). $$ g(z) = zf(z) - f(z) = a_N z^{N+1} + (a_{N-1} - a_N) z^{N} + (a_{N-2} - a_{N-1}) z^{N-1} + \cdots + (a_0 - a_1) z - a_0 $$

Since $g(z) = (z-1)f(z)$, strictly inside the unit disk the zeros of $g$ are the same as the zeros of $f$. On the boundary $g$ has a zero at $z = 1$, which needs to be handled separately.

$$ |g(z)| \ge \Big| |a_0| - |a_N z^{N+1} + (a_{N-1} - a_N) z^{N} + (a_{N-2} - a_{N-1}) z^{N-1} + \cdots + (a_0 - a_1) z | \Big| $$ $$ |g(z)| \ge \Big| |a_0| - |a_N + (a_{N-1} - a_N) + (a_{N-2} - a_{N-1}) + \cdots + (a_0 - a_1) | \Big| $$ $$ |g(z)| \ge \Big| |a_0| - |a_0 | \Big| = 0 $$ Strictly inside the unit disk there is never equality. When $|z| = 1$ you get equality when $z, z^2, \cdots z^{N+1}$ are collinear, which only happens at $z=1$. But $z = 1$ is not a zero of $f$.