[proofplan] We proceed by induction on the number of indeterminates $n$. The base case $n = 1$ is precisely [Gauss's Theorem — Polynomial Rings over UFDs are UFDs](/theorems/3245), which asserts that $R[x]$ is a UFD whenever $R$ is a UFD. For the inductive step, we write $R[x_1, \ldots, x_{k+1}] = R[x_1, \ldots, x_k][x_{k+1}]$ and apply Gauss's Theorem to the UFD $R[x_1, \ldots, x_k]$. [/proofplan]

[step:Establish the base case $n = 1$ via Gauss's Theorem]Since $R$ is a UFD, [Gauss's Theorem — Polynomial Rings over UFDs are UFDs](/theorems/3245) gives that $R[x_1]$ is a UFD. This settles the base case.[/step]

[guided]The base case asks: is $R[x_1]$ a UFD? This is exactly the content of [Gauss's Theorem — Polynomial Rings over UFDs are UFDs](/theorems/3245), which states that if $S$ is a unique factorisation domain, then the polynomial ring $S[x]$ is also a unique factorisation domain. We verify the hypothesis: Gauss's Theorem requires that the coefficient ring be a UFD. Here the coefficient ring is $R$, which is a UFD by assumption. Therefore the theorem applies, and we conclude that $R[x_1]$ is a UFD. Concretely, $R[x_1]$ inherits both the existence and uniqueness of factorisation from Gauss's Theorem: every nonzero non-unit polynomial in $R[x_1]$ factors into irreducibles (using the content-primitive decomposition and factorisation in the field of fractions), and this factorisation is unique up to reordering and multiplication by units of $R$. Why isolate the base case as a separate step rather than folding it into the induction? Because the entire multi-variable argument ultimately rests on this single application of Gauss's Theorem — the inductive step will invoke the same theorem repeatedly, each time reducing the problem to the case that has already been settled. Making the base case explicit clarifies that the one-variable result is the engine driving the whole proof. This completes the base case $n = 1$.[/guided]

[step:Carry out the inductive step by identifying $R[x_1, \ldots, x_{k+1}]$ with $R[x_1, \ldots, x_k][x_{k+1}]$] Let $k \geq 1$ and suppose, as the inductive hypothesis, that $R[x_1, \ldots, x_k]$ is a UFD. The polynomial ring in $k + 1$ indeterminates satisfies the canonical ring isomorphism \begin{align*} R[x_1, \ldots, x_{k+1}] \cong R[x_1, \ldots, x_k][x_{k+1}], \end{align*} which identifies a polynomial in $x_1, \ldots, x_{k+1}$ with a polynomial in $x_{k+1}$ whose coefficients lie in $R[x_1, \ldots, x_k]$. Since $R[x_1, \ldots, x_k]$ is a UFD by the inductive hypothesis, we apply [Gauss's Theorem — Polynomial Rings over UFDs are UFDs](/theorems/3245) to the UFD $S := R[x_1, \ldots, x_k]$ and conclude that $S[x_{k+1}] = R[x_1, \ldots, x_{k+1}]$ is a UFD.[/step]

[guided]The key algebraic observation is that the polynomial ring construction is associative: forming a polynomial ring in $k + 1$ variables at once is the same as first forming a polynomial ring in $k$ variables, then adjoining one more variable. Concretely, \begin{align*} R[x_1, \ldots, x_{k+1}] \cong R[x_1, \ldots, x_k][x_{k+1}]. \end{align*} This is a standard ring isomorphism: an element of the LHS is a finite sum $\sum_{\alpha} r_\alpha \, x_1^{\alpha_1} \cdots x_{k+1}^{\alpha_{k+1}}$, which can be regrouped by powers of $x_{k+1}$ as $\sum_{j=0}^{d} f_j \, x_{k+1}^j$ where each $f_j = \sum_{\beta} r_{\beta, j} \, x_1^{\beta_1} \cdots x_k^{\beta_k} \in R[x_1, \ldots, x_k]$. Now we apply [Gauss's Theorem — Polynomial Rings over UFDs are UFDs](/theorems/3245). The theorem requires that the coefficient ring be a UFD. By the inductive hypothesis, $R[x_1, \ldots, x_k]$ is a UFD. Therefore $R[x_1, \ldots, x_k][x_{k+1}]$ is a UFD. Under the identification above, $R[x_1, \ldots, x_{k+1}]$ is a UFD, completing the inductive step. Why does induction work so cleanly here? Because Gauss's Theorem reduces the multi-variable problem to a single-variable one: each new variable is adjoined to a ring that is already known to be a UFD. The entire argument rests on the single-variable case, applied repeatedly.[/guided]

[step:Conclude by induction that $R[x_1, \ldots, x_n]$ is a UFD for every $n \geq 1$] By the principle of mathematical induction, $R[x_1, \ldots, x_n]$ is a unique factorisation domain for every integer $n \geq 1$. [/step]