Local representation $f(z_{0})=w_{0}$ and the open mapping theorem Let $f(z)$ be a non-constant analytic function on a domain $D$ with $f(z_{0})=w_{0}$. I want to show that $f(z)-w_{0} = (z-z_{0})^{n}k(z) = g((z − z_{0})^{n}) = (h(z − z_{0}))^{n}$ with $k,g,h$ analytic and $k(z_{0})\neq{0}, g'(z_{0})\neq{0}$ and $h'(z_{0})\neq{0}$ (where $n$ is the order of the zero at $w_{0}$). Then I need to use these to prove the open mapping theorem.
At this point I have as tools the Cauchy integral formulas, Morera's and Goursat's theorem, results about power series, the power series representation of holomorphic functions and the fact that an analytic function that is not identically zero, then the zeros of the function are isolated. I do not have the maximum principle yet.
Showing $f(z) − w_{0} = (z-z_{0})^{n}k(z)$ is straightforward but I'm having trouble getting the other representations. Also I tried to prove the other part assuming the results but couldn't really get started.
Also on a non-related subject: I believe that there is a theorem that states if $f'(z_{0})={0}$, then $f$ is not one to one at that particular point. Why is this so and where can I find a reference for this (is this obtained by considering the behavior of the function $|f_{}(z)|$? )
 A: Let's take $z_0=w_0=0$ to simplify writing. This merely shifts the picture. The required representations are
$$f(z)  = z^n k(z) = g(z^n) = (h(z))^{n}$$
You have the first representation already. 
Second representation
is in general impossible to achieve.  A necessary condition for function $f$ to be written as $g(z^n)$ in a neighborhood of $0$ is: $$f(ze^{2\pi i/n})=f(z)\tag{1}$$  This is  because both $z$ and   $ze^{2\pi i/n}$ give the same value when plugged into $g(z^n)$.  
The condition (1) is also sufficient. Indeed, writing $f(z)=\sum c_k z^k$ we see from (1) that  $c^ke^{2\pi ik/n}=c_k$. Hence, $c_k=0$ unless $n$ divides $k$. Now you can write 
$$f(z) = \sum_j c_{nj}z^{nj} =g(z^n)$$
where $g(w)=\sum_j c_{nj} w^j$. 
Third representation
can be obtained from the first. You should know  something about the principal branch of the root function $z^{1/n}$: namely, that such a thing is defined on $\mathbb C\setminus (-\infty,0]$ by $re^{i\theta}\mapsto r^{1/n}e^{i\theta /n}$. 
Write $f$ as $k(0)z^n (k(z)/k(0))$. 
Pick $r>0$ such that $|k(z)/k(0) -1 |<1$   when $|z|<r$. This guarantees that $k(z)/k(0)$ is in the domain of the principal branch of the root function. Use the principal branch to define  $\psi(z)=(k(z)/k(0))^{1/n}$. Now you can write 
$$f(z) = (c z \psi(z))^n$$ where $c$ is any number such that $c^n=k(0)$. 
Open mapping theorem
Here is an approach using the first representation. Fix $r>0$ such that $k(z)\ne 0$ for  $|z|\le r$. 
We have
$$\begin{split}\int_{|z|=r} \frac{f'(z)}{f(z)}\,dz 
&=\int_{|z|=r} \frac{nz^{n-1} k(z) + z^n k'(z) }{z^nk(z)}\,dz
\\&=n \int_{|z|=r} \frac{1}{z}\,dz+ n \int_{|z|=r} \frac{k'(z)}{k(z)}\,dz
\\&=2\pi in \end{split}\tag{2}$$
where $\int_{|z|=r} \frac{k'(z)}{k(z)}\,dz=0$ because $k'/k$ is holomorphic in $|z|\le r$. Suppose there is a sequence $w_n\to 0$
such that $f$ does not attain the value $w_n$ in $|z|\le r$. Then 
$$\int_{|z|=r} \frac{f'(z)}{f(z)-w_n}\,dz =0 \tag{3}$$ for each $n$. But as $n\to\infty$, the integral (3) converges to integral (2), because
the integrand converges uniformly on $|z|=r$. Contradiction.
