You define the ''winding number about 0'' $n(f)$ of a path $f$ in the complex plane given by $f:[0,1]\to C$ where $f(0)=f(1)$ and $f$ is continuous. The way we define the winding number is intuitively, ''the number of times the curve $f$ winds around 0''. So, for instance, $n(e^{2\pi it})=1$, because this is a circle going around 0 once. $n(e^{4\pi it})=2$ because this is a circle going around 0 twice. Anyway, you can define $n(f)$ for any path $f$, although it is more tricky to do so for more complicated paths, the important fact about $n(f)$ is that it is integer valued.

Now, let $p(z)$ be a non constant polynomial in $z$, and assume $p(z)$ is never 0 (setting up for proof by contradiction). Let $f(t)=p(r{e}^{2\pi it})$, i.e. the path formed by taking the image of the circle of radius $R$ around 0. If $R=0$, this path is $f(t)=p(0)=x$ say, with $x$ not zero.

Now, whatever $R$ we take, $f(t)$ is never zero as $p(z)$ is never 0. This means if we look at the family of paths we get by varying $R$, they all lie in $C\setminus \{0\}$, i.e. the complex plane without 0.

Unfortunately at this point I have to bring in another intuitive idea as the machinery required would be too involved to go into here. As we continuously vary paths in $C\setminus \{0\}$, the winding number can't change at any point, we can see this geometrically because to increase the number of times it loops round 0, you'd have to pass through 0, which you can't do because 0 is not in the set. Maybe that's not quite as intuitively satisfying as it should be, but you can take it as true that in $C\setminus \{0\}$, what is called ''homotopic transformations'' (which basically means continuously changing one path to another) leave the winding number unchanged.

Given that we're done, because the winding number of $f$ if $R=0$ is just the winding number of the constant path $x$, which is 0. Therefore, for any $R$ the winding number of $f$ must be 0, becausewe're continuously transforming the paths in $C\setminus \{0\}$. However, for very large $R$, $p(z)$ is approximately $k{z}^n$ for some complex number $k$ not equal to zero and some integer $n\ge 1$ [because $z^n$ ismuch bigger than $z^{n-1}$ if $\left|z\right|$ is very large]. So, $f(z)$ is approximately $k{e}^{2n\pi it}$, which has winding number $n$, which is $\ge 1$.

This is a contradiction, so the assumption that $p(z)$ is never 0 must be false. Tada! The fundamental theorem of algebra.