As the title says, I'm trying to prove that $k[[X_1,\ldots,X_n]]$ is UFD for $k$ field. In Lang's Algebra, there is a proof by induction on $n$. The base of induction is clear (we even have discrete valuation ring).
Let $R_n = k[[X_1,\ldots,X_n]] \cong R_{n-1}[[X_n]]$ and assume that $R_{n-1}$ is UFD. Let $f\in R_n$, $f(X_1,\ldots,X_n)\neq 0$. Lang now handwaves that one could apply some change of variables to conclude that without loss of generality $f(0,\ldots,0,X_n)\neq 0$. Since $R_{n-1}$ is complete local ring with unique maximal ideal ${\frak m}=(X_1,\ldots,X_{n-1})$, this means that coefficients of $f$ are not all in $\frak m$. This is necessary to apply Weierstrass preparation theorem to conclude that $f = gu$ where $g\in R_{n-1}[X_n]$ and $u$ is invertible in $R_n$. Since $R_{n-1}[X_n]$ is UFD, $g$ can be factored into irreducible polynomials $g = g_1\ldots g_k$. Lang now claims that this gives existence of factorization in $R_{n-1}[[X_n]]$.
Question. Why does factorization into irreducibles of $g$ in $R_{n-1}[X_n]$ induce factorization into irreducibles in $R_{n-1}[[X_n]]$? How do we know that $g_i$'s are irreducible in $R_{n-1}[[X_n]]$ (actually they could as well be invertible, which is definitely not a problem, but what in the case that they are not?) I'm probably missing something very obvious here.
Bonus question. Is there a better way to prove the theorem?