[proofplan] We identify $R[[T_1, \dots, T_n]]$ with the $\mathfrak{m}$-adic completion of the polynomial ring $R[T_1, \dots, T_n]$, where $\mathfrak{m} = (T_1, \dots, T_n)$. Since $R$ is noetherian, [Hilbert's Basis Theorem](/theorems/2904) gives that $R[T_1, \dots, T_n]$ is noetherian. The [Properties of Noetherian Completions](/theorems/2888) then imply that the $\mathfrak{m}$-adic completion is noetherian. [/proofplan] [step:Identify $R[[T_1, \dots, T_n]]$ as the $\mathfrak{m}$-adic completion of $R[T_1, \dots, T_n]$] Let $S = R[T_1, \dots, T_n]$ and let $\mathfrak{m} = (T_1, \dots, T_n) \trianglelefteq S$ be the ideal generated by the indeterminates. Define the $\mathfrak{m}$-adic completion of $S$ as the inverse limit \begin{align*} \hat{S} := \varprojlim_{i} S / \mathfrak{m}^i. \end{align*} An element of $\hat{S}$ is a coherent sequence $(f_i + \mathfrak{m}^i)_{i \geq 0}$ with $f_i \in S$ and $f_{i+1} \equiv f_i \pmod{\mathfrak{m}^i}$ for all $i \geq 0$. Such a coherent sequence determines and is determined by a formal power series $\sum_{\alpha \in \mathbb{N}_0^n} c_\alpha T_1^{\alpha_1} \cdots T_n^{\alpha_n}$ with $c_\alpha \in R$: the truncation to terms of total degree less than $i$ gives the representative $f_i \bmod \mathfrak{m}^i$. Conversely, every formal power series defines a coherent sequence in the inverse system. This bijection is a ring isomorphism $\hat{S} \cong R[[T_1, \dots, T_n]]$. [guided] Why is the $\mathfrak{m}$-adic completion the correct object? The ideal $\mathfrak{m} = (T_1, \dots, T_n)$ consists of all polynomials with zero constant term, and $\mathfrak{m}^i$ consists of all polynomials whose terms all have total degree at least $i$. Thus $S / \mathfrak{m}^i$ is the ring of "polynomials truncated at total degree $i$," and a coherent sequence in $\varprojlim S / \mathfrak{m}^i$ is a compatible family of such truncations — which is precisely a formal power series. More precisely, define the map \begin{align*} \Phi: R[[T_1, \dots, T_n]] &\to \varprojlim_{i} S / \mathfrak{m}^i \\ \sum_{\alpha} c_\alpha T^\alpha &\mapsto \left( \sum_{|\alpha| < i} c_\alpha T^\alpha + \mathfrak{m}^i \right)_{i \geq 0}. \end{align*} This map is a ring homomorphism. It is injective because if two formal power series agree modulo $\mathfrak{m}^i$ for every $i$, they agree at every coefficient. It is surjective because any coherent sequence $(f_i + \mathfrak{m}^i)_{i \geq 0}$ determines the coefficients of a unique formal power series (the coefficient of $T^\alpha$ is determined once $i > |\alpha|$). Hence $\Phi$ is a ring isomorphism. [/guided] [/step] [step:Apply Hilbert's Basis Theorem to conclude $S = R[T_1, \dots, T_n]$ is noetherian] Since $R$ is noetherian, [Hilbert's Basis Theorem](/theorems/2904) asserts that every finitely generated algebra over a noetherian ring is noetherian. The polynomial ring $S = R[T_1, \dots, T_n]$ is a finitely generated $R$-algebra (generated by $T_1, \dots, T_n$), so $S$ is noetherian. [/step] [step:Apply the noetherianness of completions to conclude $R[[T_1, \dots, T_n]]$ is noetherian] By the [Properties of Noetherian Completions](/theorems/2888) (part 1), if $S$ is a noetherian ring and $\mathfrak{m} \trianglelefteq S$ is an ideal, then the $\mathfrak{m}$-adic completion $\hat{S}$ is noetherian. We have verified that $S = R[T_1, \dots, T_n]$ is noetherian and $\mathfrak{m} = (T_1, \dots, T_n) \trianglelefteq S$ is an ideal. Therefore $\hat{S}$ is noetherian. Since $\hat{S} \cong R[[T_1, \dots, T_n]]$ by the identification in the first step, the ring $R[[T_1, \dots, T_n]]$ is noetherian. [guided] The structure of this proof is a composition of two major results: 1. [Hilbert's Basis Theorem](/theorems/2904) ensures the polynomial ring $R[T_1, \dots, T_n]$ inherits noetherianness from $R$. 2. The [Properties of Noetherian Completions](/theorems/2888) ensures the $\mathfrak{m}$-adic completion of a noetherian ring (with respect to any ideal) is again noetherian. The only substantive work in this proof is the identification of $R[[T_1, \dots, T_n]]$ with the $\mathfrak{m}$-adic completion of $R[T_1, \dots, T_n]$ at the ideal $\mathfrak{m} = (T_1, \dots, T_n)$. Once this identification is established, the noetherianness of $R[[T_1, \dots, T_n]]$ is an immediate consequence. Note that this argument handles multiple variables at once: we do not need to induct on $n$ (as one would if proving the result directly), because [Hilbert's Basis Theorem](/theorems/2904) already handles the polynomial ring in $n$ variables, and the completion result applies to any ideal in any noetherian ring. [/guided] [/step]