Denote by $R_{n}=\mathbb{C}\{z_{1},\cdots, z_{n}\}$ the ring of convergent power series. We say that
$f\in \mathbb{C}\{z_{1},\cdots, z_{n}\}$ is $z_n$-general of order $m$ if there is an $h\in\mathbb{C}\{z_n\}$, $h(0)\not=0$ such that $f(0,\cdots,0,z_{n})=z^{m}_{n}h(z_n)$.
A polynomial $\omega = z_{n}^{m} + a_{m-1}z_{n}^{m-1} + \cdots + a_{0} \in R_{n-1}[z_n]$ which satisfies $a_{j}(0)=0$ for all $j=0,\cdots, m-1$ is called a Weierstrass polynomial of degree $m$.
Also, we have the following results:
Weierstrass Division Theorem : For each $f\in R_{n}$ be $z_n$-general of order $m$ and $g\in R_{n}$. Then there is an unique power series $q\in R_{n}$ and an unique polynomial $r\in R_{n-1}[z_{n}]$ of degree $<m$ such that $g=qf+r$.
Weierstrass Preparation Theorem : For each $f\in R_{n}$ be $z_n$-general of order $m\in\mathbb{N}$. Then there are an unique Weierstrass polynomial $\omega\in R_{n-1}[z_n]$ and an unique unit $u\in R_{n}$ such that $f=u\cdot \omega$.
I have managed to show that Weierstrass Division Theorem implies Weierstrass Preparation Theorem, using $g$ as the polynomial $z_{n}^{m}$ then there are $q\in R_{n}$ and $r\in R_{n-1}[z_n]$ of degree $<m$. It follows that $q$ is a unit and $z^{m}_{n}-r$ is a Weierstrass polynomial.
Does anyone have any idea how to prove that Weierstrass Preparation Theorem implies Weierstrass Division Theorem?