The truncated polynomial ring $k[x]\mathrm{/}(x^n)$ is local Artinian of length $n$

By For every field $F$, $F[x]$ is a principal ideal domain, every ideal of $k[x]$ is principal. The ideals of $R$ correspond by Correspondence theorem: ideals of $R\mathrm{/}I$ correspond to ideals of $R$ containing $I$ to the ideals of $k[x]$ containing $(x^n)$, hence to the principal ideals $(f)$ with $f$ dividing $x^n$. Up to multiplication by a unit, these are exactly $(x^r)$ for $0\le r\le n$. Therefore the ideals of $R$ are precisely $R=(x^0)⊋(x)⊋\cdots ⊋(x^{n-1})⊋(x^n)=0$, so $(x)$ is the unique maximal ideal.

The chain in step 1.1 shows directly that $R$ is Artinian and that $(x{)}^n=0$. For each $0\le r<n$, the quotient $(x^r)\mathrm{/}(x^{r+1})$ is generated by the class of $x^r$ and annihilated by $x$, so it is one-dimensional over the residue field $R\mathrm{/}(x)\cong k$. Hence each quotient has length $1$.

Applying Module length is additive in short exact sequences successively to $0\to (x^{r+1})\to (x^r)\to (x^r)\mathrm{/}(x^{r+1})\to 0$ for $0\le r<n$ shows that ${\mathrm{\ell }}_R(R)=n$. Thus $R$ is a local Artinian ring of length $n$.