Hilbert's Basis Theorem

Step 1 (leading-coefficient ideals). Let $I \subseteq R[x]$ be an ideal. For each degree $n \ge 0$, let $L_n$ be the set of leading coefficients of degree-$n$ elements of $I$, together with $0$. Multiplying a degree-$n$ polynomial by $x$ shows $L_n \subseteq L_{n+1}$, and each $L_n$ is an ideal of $R$ (closure under addition and absorption of $R$-multiples follows directly from $I$ being an ideal).

Step 2 (use that $R$ is Noetherian, twice). The chain $L_0 \subseteq L_1 \subseteq L_2 \subseteq \cdots$ stabilizes at some $L_N$ by the ACC. Since $R$ is Noetherian, each of $L_0,\ldots,L_N$ is finitely generated; choose finitely many generators of each $L_n$ ($n=0,\ldots,N$) and, for each generator, a polynomial $f_{n,i} \in I$ of degree $n$ realizing it as leading coefficient. This gives one finite list $\{f_{n,i}\}$ in total.

Step 3 (reduction by induction on degree). We claim $I$ is generated by this finite list. Take any $f \in I$ of degree $d$; we induct on $d$. If $d \le N$, the leading coefficient of $f$ lies in $L_d$, so it is an $R$-combination of the leading coefficients of the $f_{d,i}$; subtracting the matching $R$-combination of the $f_{d,i}$ from $f$ cancels the degree-$d$ term, producing an element of $I$ of strictly smaller degree, and we induct. If $d > N$, the leading coefficient of $f$ lies in $L_d = L_N$, so it is an $R$-combination of the leading coefficients of the $f_{N,i}$; subtracting the same combination of $x^{\,d-N} f_{N,i}$ again cancels the leading term and strictly lowers the degree.

Step 4 (conclusion). Repeating Step 3 eventually produces the zero polynomial, so $f$ is an $R[x]$-combination of the finite list $\{f_{n,i}\}$. Hence every ideal of $R[x]$ is finitely generated, i.e. $R[x]$ is Noetherian. Applying this inductively $n$ times starting from the Noetherian ring $k$ shows $k[x_1,\ldots,x_n] = k[x_1][x_2]\cdots[x_n]$ is Noetherian.