In a proof of a theorem in chapter 2 "p-adic equations" in "A Course in Arithmetic" from J-P Serre there is one conclusion that I don't understand. Here is the theorem I'm talking about (excluding the lemma it is using, since I don't think it is important for my questions)
Theorem 1: Let $f \in \mathbb{Z}_p[X_1,\ldots,X_m], x = (x_i) \in (\mathbb{Z_p})^m,n,k \in \mathbb{Z}$ and $j$ an integer such that $1 \leq j \leq m$. Suppose that $0 \leq 2k < n$ and that
\begin{equation} f(x) \equiv 0 \mod p^n \text{ and } \nu_p \left( \frac{\partial f}{\partial X_j}(x)\right) = k.\end{equation}Then there exists a zero $y$ of $f$ in $(\mathbb{Z}_p)^m$ which is congruent to $x$ modulo $p^{n-k}$.
Proof: Suppose first that $m=1$. By applying the above lemma to $x^{(0)} = x$, we obtain $x^{(1)} \in \mathbb{Z}_p$ congruent to $x^{(0)} \mod p^{n-k}$ and such that \begin{equation} f(x^{(1)}) \equiv 0 \mod p^{n+1} \text{ and } \nu_p(f'(x^{(1)})) = k. \end{equation} Arguing inductively, we construct in this way a sequence $x^{(0)}, \ldots, x^{(q)}, \ldots$ such that \begin{equation} x^{(q+1)} \equiv x^{(q)} \mod p^{n+q-k}, f(x^{(q+1)}) \equiv 0 \mod p^{n+1}. \end{equation}
This is a Cauchy sequence. If $y$ is its limit, we have $f(y) = 0$ and $y \equiv x \mod p^{n-k}$, hence the theorem for $m=1$.
My questions are, why does $f(y) = 0$ and $y \equiv x \mod p^{n-k}$ hold?
Using the distance defined as $d(x,y) = \exp(-\nu_p(x-y))$, where $\nu_p(x)$ is the p-adic valuation.
What I have is, that $d(f(x^{(q)}),0)$ tends to zero for $q \to \infty$ by doing the following:
Since $f(x^{(q)}) \equiv 0 \mod p^{n+q}$, we have $\nu_p(f(x^{(q)})) \geq n+q$. Therefore: $d(f(x^{(q)}),0) = \exp(-\nu_p(f(x^{(q)}))) \leq \frac{1}{\exp (n+q)} \to 0$ for $q \to \infty$.
How do I advance from here? It would be nice to have some continuity, but since Serre didn't introduce it before, I don't think, this is the way to proof this...
And I have no clue how I can prove the second conclusion ($y \equiv 0 \mod p^{n-k}$). I know, that every $x^{(q)}$ fulfills that relation, but that doesn't need to be true in the limit case (a priori).