Proof that $A[[X_1,\ldots,X_r]]$ is complete in the $(X_1,\ldots,X_r)$-adic topology. I just want to check whether the following argument is correct. Thank you!

Let $B=A[[X_1,\ldots,X_r]]$ where $A$ is a commutative ring and let $I=(X_1,\ldots,X_r).$
Let $(f_m)$ be a Cauchy sequence in $B.$
We can find an increasing sequence of positive integers $N(1)\leq N(2)\leq \ldots$ such that
$$m\geq N(n)\; \Rightarrow \;f_m-f_{N(n)}\in I^n.$$
For $n \in \mathbb{N}$ and $g \in B,$ we define $P_n(g)$ to be the degree $n$ homomogeneous polynomial part of $g,$
e.g. $P_3(1+X_1^2+X_1X_r^2+X_r^3+\text{terms of higher degree})=X_1X_r^2+X_r^3.$
We now define $f=P_0(f_{N(1)})+P_1(f_{N(2)})+\ldots \in B.$
Claim: $f_m \to f$ in the $I$-adic topology on $B.$
Proof: Let $n \in \mathbb{N}$ be given, we must find $m \in \mathbb{N}$ such that $f_m - f \in I^n.$
Observe that $f_m-f \in I^n$ if and only if $P_r(f_m-f)=0$ for all $0\leq r <n.$
Also we have $P_r(f_m-f)=P_r(f_m-f_{N(r)}).$
Hence $f_m-f \in I^n$ if and only if $P_r(f_m-f_{N(r+1)})=0$ for all $0\leq r <n.$
Thus $f_m-f \in I^n$ if and only if $f_m-f_{N(r+1)} \in I^{r+1}$ for all $0\leq r <n.$
We can therefore take $m=N(n). \; \; \; $ Q.E.D.
 A: The proof is correct, but it should read backwards from a certain point of view, which is unpleasant for the reader. Let me use the OP's idea to give (almost) the same proof in slightly different (and many quite some more) words. Differences in notation are just a matter of taste.
So, we have the ring of formal power series $A = k[[x_1,x_2,\dots,x_r]]$ and consider the (maximal) ideal $\mathfrak{m} = (x_1,x_2,\dots,x_r)$; we aim to show that every Cauchy sequence in $A$ with respect to the $\mathfrak{m}$-adic topology converges in the $\mathfrak{m}$-adic topology.
For late use, we introduce the following notation: Given $f = \sum f_Ix^I\in A$, we let $$\tau^{\leq d}(f) = \tau^{< d+1}(f) := \sum_{\# I\leq d}f_Ix^I$$ be the truncation of $f$ at degree $d$. Then the term in degree $d$ is precisely $\tau^{\leq d}(f)-\tau^{<d}(f)$.
The key observation for the proof is that for each $d\in\mathbb{N}$ we have $f-g\in\mathfrak{m}^{d}$ if and only if $\tau^{< d}(f) = \tau^{< d}(g)$.
Let $(f_m)_{m\in \mathbb{N}}$ be a Cauchy sequence in $A$; by definition of the topology, this boils down to say that for each $n\in \mathbb{N}$ there exists an $N\in\mathbb{N}$ such that $f_m-f_{m'}\in\mathfrak{m}^n$ for all $m,m'\geq N$.
The key observation translates this into $\tau^{<n}(f_m) = \tau^{<n}(f_{m'})$ for all $m,m'\geq N$.
On the other hand, the sequence $(f_m)_{m\in\mathbb{N}}$ converges to an element $f$ if for each $n\in \mathbb{N}$ there exists an $N\in\mathbb{N}$ such that $f_{m}-f\in \mathfrak{m}^n$ for all $m\geq N$; i.e., if $\tau^{<n}(f) = \tau^{<n}(f_m)$ for all $m\geq N$.
We will show that this happens for the following choice of $f$:
For each $n\in \mathbb{N}$ we pick an $N(n)\geq N(n-1)$ such that $f_m-f_{N(n)}\in \mathfrak{m}^n$ for all $m\geq N(n)$ and we define the $(n-1)$-th term of $f$ to be the $(n-1)$-th term of $f_{N(n)}$.
That is, $f\in A$ is the unique element such that
$$\tau^{< n}(f)-\tau^{<n-1}(f)=\tau^{< n}(f_{N(n)})-\tau^{<n-1}(f_{N(n)})$$
for each $n\geq 2$ and $\tau^{<1}(f) = \tau^{<1}(f_{N(1)})$.
In particular,
$$
\begin{align}\tau^{< n}(f)-\tau^{< n}(f_{N(n)}) &= \tau^{<n-1}(f)-\tau^{< n-1}(f_{N(n)})\\
&=\tau^{<n-1}(f)-\tau^{<n-1}(f_{N(n-1)}),\end{align}$$
where $\tau^{<n-1}(f_{N(n)}) = \tau^{<n-1}(f_{N(n-1)})$ since $N(n)\geq N(n-1)$ and so $f_{N(n+1)}-f_{N(n)}\in\mathfrak{m}^{n}$.
Inductively, this shows that $\tau^{\leq n}(f)-\tau^{\leq n}(f_{N(n+1)}) =\tau^{<1}(f)-\tau^{<1}(f_{N(1)})=0$ and, in particular, $f-f_{N(n)}\in\mathfrak{m}^n$ for all $n\in\mathbb{N}$.
As we chose $N(n)$ such that $f_m-f_{N(n)}\in\mathfrak{m}^n$ for all $m\geq N(n)$ this finally shows that
$$f-f_m = (f-f_{N(n)})+(f_{N(n)}-f_m)\in\mathfrak{m}^n$$
for all $m\geq N(n)$ and so the sequence $(f_m)_{m\in \mathbb{N}}$ indeed converges to $f$ in the $\mathfrak{m}$-adic topology. 
