p-adic arithmetic, Cauchy sequence, polynomial I have a couple questions on the following theorem and proof, taken from here: https://www.math.purdue.edu/~lipman/MA598/Serre-Course%20in%20Arithmetic.pdf
Let $f\in\mathbb{Z}_p[X_1,\dotso, X_n], x=(x_i)\in (\mathbb{Z}_p)^m, n, k\in\mathbb{Z}$ and $j$ an integer such that $0\leq j\leq m$. Suppose that $0<2k<n$ and that $f(x)\equiv 0\mod p^n$ and $v_p(\frac{\partial f}{\partial X_j}(x))=k$.
Then there exists a zero $y$ of $f$ in $(\mathbb{Z}_p)^m$ which is congruent to $x\mod p^{n-k}$.
Proof:
Suppose first that $m=1$. By applying Hensels lemma (p-adic analogue of Newton's method) to $x^{(0)}=x$, we obtain $x^{(1)}\in\mathbb{Z}_p$ congruent to $x^{(0)}\mod p^{n-k}$ and such that $f(x^{(1)})\equiv 0\mod p^{n-k}$ and $v_p(f'(x^{(1)}))=k$.
We can apply the lemma to $x^{(1)}$, after replacing $n$ by $n+1$. Arguing inductivly, we construct in this way a sequence $x^{(0)},\dotso, x^{(q)},\dotso$ such that $x^{(q+1)}\equiv x^{(q)}\mod p^{n+q-k}$, $f(x^{(q)})\equiv 0\mod p^{n+q}$.
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 $m=1$.
The case $m>1$ reduces to the case $m=1$ by modifying only $x_j$. 
More precisly, let $\hat{f}\in\mathbb{Z}_p[X_j]$ be the polynomial in one variable obtained by replacing $X_i$, $i\neq j$ by $x_i$.
What has just been proven can be applied to $\hat{f}$ and $x_j$.
This shows the existence of $y\equiv x_j\mod p^{n-k}$ such that $\hat{f}(y_i)=0$.
If one puts $y_i=x_i$ for $i\neq j$, the element $y=(y_i)$ satisfies the desired condition.

Questions:
1) I would like to give the construction of the sequence with less "explaining" and work out the inductive part. How could I do that?
2) How do I show, that the sequence is a Cauchy sequence?
3) In the case $m>1$, why can we simply replace $X_i$ by $x_i$? 

My thoughts on 2) and 3):
ad 3):
Is it, that $\mathbb{Z}_p[X_1,\dotso X_n]$ is factoriel, just like $\mathbb{Z}[X_1,\dotso, X_n]$, and because of that we can replace it like in the proof?
In that case, how can we see, that $\mathbb{Z}_p[X_1,\dotso X_n]$ is factoriel?
ad 2):
The metric is given by $d(x,y)=e^{-v_p(x+y)}$.
I have to show, that for every $\epsilon >0$ exists $N\in\mathbb{N}$ such that for every $r,t\in\mathbb{N}$ holds, that $d(x^{(r)}, x^{(t)})<\epsilon$.
I struggle in giving an exact proof.
$d(x^{(r)}, x^{(t)})=e^{-v_p(x^{(r)}-x^{(t)})}\leq e^{\inf(v_p(x^{(r)}), v_p(x^{(t)}))}\leq e^{-v_p(x^{(r)})}$, 
where $\inf(v_p(x^{(r)}), v_p(x^{(t)}))=v_p(x^{(r)})$
How can we proceed from here?
Thanks in advance.
 A: 1) The detailed inductive step will proceed as follows: suppose we have found $x^{(q)}\equiv x\pmod{p^{n-k}}$ with $f(x^{(q)})\equiv 0\pmod{p^{n+q}}$ and $v_p(f'(x^{(q)}))=k$. We apply the lemma with $n+q$ in place of $n$, to find $y$ satisfying $y\equiv x^{(q)}\pmod{p^{n+q-k}}$ (hence $y\equiv x\pmod{p^{n-k}}$), $f(y)\equiv 0\pmod{p^{n+q+1}}$ and $v_p(f'(y))=k$. Let $x^{(q+1)}=y$.
2) Note that, for $r\geq t$, we have $x^{(r)}\equiv x^{(t)}\pmod{p^{n+t-k}}$ by construction. Hence $v_p(x^{(r)}-x^{(t)})\geq n+t-k$, and it follows that, for any $r,t$,
$$d(x^{(r)},x^{(t)})=e^{-v_p(x^{(r)}-x^{(t)})}\leq e^{-(n-k+\min(r,t))},$$
and now it's clear the sequence is Cauchy.
3) We want to prove the case of arbitrary $m$ by reducing to the case $m=1$. To do this, we first find a new polynomial $\hat f$ with just one variable to which we apply that case. We take
$$\hat f(X) = f(x_1,\dots,x_{j-1},X,x_{j+1},\dots,x_m).$$
By assumption, $\hat f(x_j)=f(x)=0$ and $v_p\left(\hat f'(x_j)\right)=v_p\left(\frac{\partial f}{\partial X_j}(x)\right)=k$, so we can apply what was already proven to find a root $y_j\equiv x_j\pmod{p^{n-k}}$ of $\hat f$. Then
$$f(x_1,\dots,x_{j-1},y_j,x_{j+1},\dots,x_m)=0,$$
giving the desired zero of $f$.
