Proof:

If the polynomial $p(z) = z^n + a_1z^{n-1} + \dots + a_n$ has no zeros, then in particular, $p(1) \neq 0$. For $t \neq 0$, we define

$$p(z/t)t^n = \Big(z^n + a_1z^{n-1}t + \dots + a_nt^n\Big).$$

The right hand side makes sense even when $t = 0$ and we denote the right hand side by $g(z, t)$. Observe that $g(z, 0) = z^n$ and $g(z, 1) = p(z)$. However we need a homotopy of maps of $S^1$ preserving the base point $1$. To this end we modify it consider instead the map $F:S^1\times [0, 1]\longrightarrow S^1$ given by

$$F(z, t) = \frac{g(z, t)}{\vert g(z, t)\vert}\frac{\vert g(1, t)\vert}{g(1, t)}. \eqno(12.7)$$

Clearly $g(z, 0)\neq 0$ for any $z \in S^1$ and if $0 < t \leq 1$ then again $g(z, t) = p(z/t)t^n \neq 0$. Thus (12.7) is a base point preserving homotopy between the function $f : S^1 \longrightarrow S^1$ given by

$$f(z) = \frac{p(z)}{\vert p(z)\vert}\frac{\vert p(1)\vert}{p(1)} \eqno(12.8)$$

and the map $z \mapsto z^n$. We conclude that degree of $f$ is $n$. However we have a base point preserving homotopy between (12.8) and the constant map namely, $G:S^1\times [0, 1]\longrightarrow S^1$ given by

$$G(z, s) = \frac{p(sz)}{\vert p(sz)\vert}\frac{\vert p(s)\vert}{p(s)}.$$

We now conclude that degree of (12.8) is zero and we have a contradiction.