Roots of derivative of polynomial Suppose f(z) is a polynomial such that its derivatives are non-zero for all $|z| <1$.   Is the restriction of $f$ to $|z|<1$ 1 to 1?
I know that $f$ must be locally 1 to 1.   It is obvious that $f$ is 1-1 for polynomials of order 1.  The case of order 2 follows from Gauss-Lucas Thm.
I am stuck on how to prove the general case.
 A: It's not true in general... consider $f(z) = (z-1)^n - 1$. Then its derivatives have zeroes only at $z = 1$, but it takes the value zero at any $z = 1 + e^{2\pi i / n}$. Since these are spaced evenly around the circle $|z - 1|= 1$, for large $n$ many of them will lie inside the unit disk.
If you want an example where the derivatives have zeroes only outside the ${\it closed}$ unit disk, then you can take $f(z) = (z-(1+\epsilon))^n - 1$ for small $\epsilon > 0$.
A: Zarrax has already given a great example.  Here's a way to see that there are polynomial examples from the fact that there are holomorphic examples.  The motivation for this is that the non-polynomial example $g(z)=e^{2\pi i z}$ came to mind when I read your question.
Let $g$ be holomorphic in a neighborhood of the closed disk, not injective on the open disk, but with nonvanishing derivative on the closed disk.  Let $(p_n)_n$ be the sequence of partial sums of the Taylor series of $g$ centered at $0$, so $p_n\to g$ and $p_n'\to g'$ uniformly on a (smaller) neighborhood of the closed disk.  
Since $g'$ is nonvanishing on the closed disk, it has a positive minimum modulus $m$ there, which by uniform convergence means that $p_n'$ eventually has modulus greater than $m/2$, and in particular is eventually nonvanishing on the closed disk.  Take $a\neq b$ in the open unit disk such that $g(a)=g(b)$.  Applying Hurwitz's theorem to the sequence of functions $p_n(z)-g(a)$ on small disjoint disks centered at $a$ and $b$ shows that $p_n$ eventually takes on the value $g(a)$ at more than one point in the open unit disk.  Thus, $p_n$ is eventually a counterexample.
In fact, using WolframAlpha, it looks like $f(z)=\displaystyle{\sum_{k=0}^{25}\frac{(2\pi iz)^k}{k!}}$ gives an example.  It allegedly takes on the value $-1$ at points very close to $\pm\frac{1}{2}$, and all $24$ of the zeros of its derivative are outside the closed disk.
