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We know the following things:

$f(z)$ is holomorphic on $\mathbb{C}$, $f(z)$ is injective and $f(z)$ is a polynomial, so $f'(z)$ is a polynomial too.

(1) Why is $f'(z) \neq 0$? Or why does $f$ have only one zero?

(2) Why is $f'(z)$ constant if (1) holds?

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Injective means that, in particular, $f(z_1) = f(z_2) =0$ implies $z_1=z_2.$ This means $f(z)=a(z-z_1)^n$, with $a\ne 0,$ $n= \deg f.$ Since $f(z_1 + \omega )=1$ where $\omega $ is an $n$th root of unity, injectivity gives once again that $n=1.$ Hence $f'(z) = a.$

What's more interesting is that you don't need to require that $f$ is a polynomial - here you actually use the fact that $f$ is entire.

EDIT: Here's a sketch.

$f$ is injective and analytic, so the open mapping theorem says that $f( B_{\epsilon }(1) )$ contains a neighborhood of $f(1).$ Define $g(z) = f(\frac{1}{z}).$ If $f$ does not have a power series expressible as a polynomial, $g$ has an essential singularity at $0.$ Since $g$ is holomorphic on $\mathbb{C} / \{ 0 \} ,$ the Casorati-Weierstrass Theorem lets us construct a sequence $\{ \zeta_n \}$ converging to $0$ such that $g( \zeta_n ) $ converges to $f(1).$ For $n$ sufficiently large, $z_n \notin B_{\epsilon }(1),$ contradicting injectivity.

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Has the fundamental theorem of algebra any role here? –  Chris Jun 17 '12 at 21:43
    
It allows us to write $f$ in the factored form, but the argument it really comes down to using the definition of injectivity. –  user17794 Jun 17 '12 at 21:48
    
To be clear: the interesting parts that don't require $f$ to be polynomial are that if $f$ is holomorphic and injective, $f'$ is never $0$, and that $f$ has exactly one zero (obviously there can't be more than one since $f$ is injective). –  Robert Israel Jun 17 '12 at 21:49
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@Robert and Tim: why does $f$ holomorphic and injective imply that $f$ has a zero? (I can prove it, but is there an elementary argument?) –  Georges Elencwajg Jun 17 '12 at 22:04
    
@Georges , Well, if you consider the open mapping theorem and the Casorati-Weierstrass Theorem elementary (I don't.) I added my argument to the original post - I think it's right. –  user17794 Jun 17 '12 at 23:28
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