I have trouble understanding the proof of Chapter IV Theorem 9.3 in Lang's Algebra, 3rd edition, where he proves that if $k$ is a field, then the ring of formal power series in $n$ variables is a UFD, using Weierstrass Preparation. Let $R_n=k[[X_1,...,X_n]]$. First he does some linear change of variables (which already seems mysterious), then write $f\in R_n$ as $f=gu$, where $u\in R_n$ is a unit and $g\in R_{n-1}[X_n]$ is the Weiertrass polynomial. Since $R_{n-1}[X_n]$ is UFD, we can write $g=g_1\cdots g_r$ irreducibles, and he says this shows existence of factorization. But the $g_1,...,g_r$ are only irreducible in $R_{n-1}[X_n]$, not necessarily in $R_n=R_{n-1}[[X_n]]$, so why are we done here? Can we instead first prove that $R_n$ is Noetherian, and then deduce existence of factorizaton?
The uniqueness part also puzzles me. Lang says let $f=f_1\cdots f_s$ be a factorizaton, then write $f_q=h_qu_q'$ for $q=1,...,s$, where $h_q\in R_{n-1}[X_n]$ is the Weierstrass polynomial and $u_q'$ is a unit. Then he claims $h_q$ must be irreducible in $R_{n-1}[X_n]$, but why? For example $X^2+(Y+1)X+Y=(X+Y)(X+1)$ is irreducible in $k[[Y,X]]$ since $X+1$ is a unit, but it's reducible in $k[[Y]][X]$.
I guess I've figured out the uniqueness part. Denote by $\mathfrak{m}$ the maximal ideal of a local ring $R$. $f\in R[X]$ is called a Weierstrass polynomial if it's of form $X^m+b_{m-1}X^{m-1}+\cdots+b_1X+b_0$, where all $b_i\in\mathfrak{m}$. We claim that if $f$ is reducible in $R[X]$, then it is also reducible in $R[[X]]$. Suppose in $R[X]$ we have $f=gh$. It suffices to show that $g,h$ cannot be units in $R[[X]]$. A formal power series is a unit iff its constant term is a unit in $R$. Suppose for contradiction that the constant term of $g$ is a unit. Write $h=c_kX^k+c_{k-1}X^{k-1}+\cdots+c_1X+c_0$. Let $i$ be the smallest index such that $c_i$ is a unit (which exists because $c_k$ is a unit). Then $b_i=c_ib_0+$ something in $\mathfrak{m}$, so it's a unit, contradicting the definition of Weierstrass polynomial.