I would like to see whether the following proof is correct or not.
Let $\Omega$ be a completion of an algebraic closure of $\mathbb{Q}_p$. Let $g$ and $g_1$ be power series in $1+X\Omega[[X]]$ that are converging and not vanishing on a closed disc $D(p^\lambda)$. Let $h$ and $h_1$ be polynomials of degree $N$ in $1+X\Omega[X]$ with all their zeroes belonging to $D(p^\lambda)$. Assume that
$$h(X)g_1(X)=h_1(X)g(X),$$
we want to show that $h=h_1$. Let us proceed by induction on $N$. Consider the case $N=1$, if $\alpha\in D(p^\lambda)$ is a root of $h$, then the following expression makes sense $$0=h(\alpha)g_1(\alpha)=h_1(\alpha)g(\alpha),$$ implying that $h_1(\alpha)=0$ and so $h=h_1$, since they have degree $1$. Assume now that $N> 1$ and the statement true for $N-1$. As before we can show that if $\alpha\in D(p^\lambda)$ is a root of $h$, then $h(\alpha)=0=h_1(\alpha)$. So there exist polynomials $h'$ and $h_1'$ in $1+X\Omega[X]$, of degree $N-1$, such that
$$h(X)=(1-\frac{X}{\alpha})h'(X) \quad \text{and} \quad h_1(X)=(1-\frac{X}{\alpha})h_1'(X),$$ giving
$$(1-\frac{X}{\alpha})h'(X)g_1(X)=(1-\frac{X}{\alpha})h_1'(X)g(X).$$
Being $\Omega[[X]]$ an integral domain we obtain $h'(X)g_1(X)=h_1'(X)g(X)$ and we conclude by inductive hypothesis.
I am asking this question because this is something that is proved in Theorem 14 ($p$-adic Weierstrass Preparation Theorem), p. 105 in Koblitz's "$p$-adic Numbers, $p$-adic Analysis and Zeta-Functions" in a different way. The author uses Lemma 8 (p. 104) to ensure some convergence. I would like to understand why this Lemma is necessary.
Edit: I am adding what is written on the book.
Lemma 8. Let $f(X) \in 1+\Omega[[X]]$ converge and have value $0$ at $\alpha$. Let $g(X)$ be obtained by dividing $f(X)$ by $(1 — X/\alpha)$, or equivalently, by multiplying $f(X)$ by the series $1+X/\alpha+X^2/\alpha^2+\dots$. Then $g(X)$ converges on $D(|\alpha|_p)$.
Then the proof of $h(X)=h_1(X)$ goes like this:
We proceed by induction on $N$ the degree of $h$. The case $N=1$ is like above. Suppose $N>1$, without loss of generality we may assume that $- \lambda$ is the $\text{ord}_p$ of a root $\alpha$ of $h(X)$ having minimal $\text{ord}_p$. Since $\alpha$ is a root of both $h(X)$ and $h_1(X)$ of minimal $\text{ord}_p$, we can divide both sides of the equality $h(X)g_1(X)=h_1(X)g(X)$ by $(1-X/\alpha)$, using Lemma 8, and thereby reducing to the case of our claim with $N$ replaced by $N-1$.