Solution:

Assume that $|z| < 1$. If $a_{0} z^{n} + a_{1} z^{n-1} + \cdots + a_{n} = 0$ then $a_{0} z^{n+1} + a_{1} z^{n} + \cdots + a_{n} z = 0$ and subtracting these two equations leads to $a_{0} z^{n+1} + (a_{1} - a_{0}) z^{n} + \cdots + (a_{n} - a_{n-1}) z - a_{n} = 0$, or equivalently $a_{n} = a_{0} z^{n+1} + (a_{1} - a_{0}) z^{n} + \cdots + (a_{n} - a_{n-1}) z$ hence

$$\begin{aligned} |a_{n}| & = \left| a_{0} z^{n+1} + (a_{1} - a_{0}) z^{n} + \cdots + (a_{n} - a_{n-1}) z \right| \\ & \leq a_{0} |z|^{n+1} + (a_{1} - a_{0}) |z|^{n} + \cdots + (a_{n} - a_{n-1}) |z| \\ & < a_{0} + (a_{1} - a_{0}) + \cdots + (a_{n} - a_{n-1}) = a_{n}, \end{aligned}$$

which is impossible. Thus $|z| \geq 1$.