An Artinian local ring has nilpotent maximal ideal, and its finite modules have finite length

Because $R$ is local, $\mathfrak{m}$ is its only maximal ideal. By Every prime ideal of an Artinian ring is maximal, every prime ideal of $R$ is maximal, so $\mathfrak{m}$ is also the only prime ideal. Therefore The nilradical is the intersection of all prime ideals gives $\mathrm{Nil}(R)=\mathfrak{m}$. Now The nilradical of an Artinian ring is a nilpotent ideal yields an integer $n\ge 1$ with ${\mathfrak{m}}^n=0$.

By Every commutative Artinian ring is Noetherian, the ring $R$ is Noetherian. Hence A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member makes each ideal ${\mathfrak{m}}^i$ finitely generated; fix generators $a_{i1},\dots ,a_{i{s}_i}$ of ${\mathfrak{m}}^i$. Also choose generators $x_1,\dots ,x_t$ of $M$. Then for every $0\le i<n$, the quotient ${\mathfrak{m}}^{i}M\mathrm{/}{\mathfrak{m}}^{i+1}M$ is spanned over the residue field $k=R\mathrm{/}\mathfrak{m}$ by the finitely many classes of the elements $a_{iq}{x}_j$. Indeed every element of ${\mathfrak{m}}^{i}M$ is a finite sum ${\sum }_q{a}_{iq}{m}_q$, and each $m_q$ is a finite $R$-linear combination of the $x_j$; modulo ${\mathfrak{m}}^{i+1}M$, only the residue classes of the coefficients in $k$ matter because multiplication by an element of $\mathfrak{m}$ lands in ${\mathfrak{m}}^{i+1}M$. Deleting redundant spanning vectors yields a basis $y_1,\dots ,y_d$ of ${\mathfrak{m}}^{i}M\mathrm{/}{\mathfrak{m}}^{i+1}M$, and the partial spans $0<k{y}_1<k{y}_1+k{y}_2<\cdots <k{y}_1+\cdots +k{y}_d$ form a composition series. So every quotient ${\mathfrak{m}}^{i}M\mathrm{/}{\mathfrak{m}}^{i+1}M$ has finite length.

The filtration $M\supseteq \mathfrak{m}M\supseteq \cdots \supseteq {\mathfrak{m}}^{n}M=0$ is finite by step 1.1. Applying Module length is additive in short exact sequences successively to $0\to {\mathfrak{m}}^{i+1}M\to {\mathfrak{m}}^{i}M\to {\mathfrak{m}}^{i}M\mathrm{/}{\mathfrak{m}}^{i+1}M\to 0$ shows that $M$ has finite length. Taking $M=R$ recovers the ring case from the first sentence of the theorem.