= Solution
A <local ring> is a nonzero ring with one <maximal ideal> $\mathfrak m$. Write $\kappa=R/\mathfrak m$ for its <residue field>. Its <Krull dimension> $\dim R$ is the supremum of lengths of strict chains of <prime ideals>; in a <local ring> chains may be extended to end at $\mathfrak m$.
For the local Hilbert-function convention, define the <Hilbert–Samuel growth dimension>
$$
\boxed{d(R)=\deg\bigl(\text{the eventual polynomial }
H_R(t)=\operatorname{length}(R/\mathfrak m^t)\bigr).}
$$
Equivalently it is the order of the pole at $z=1$ of the <Hilbert series> of the <associated graded ring>
$$
\operatorname{gr}_{\mathfrak m}R=\bigoplus_{j\geq0}\mathfrak m^j/\mathfrak m^{j+1}.
$$
This finite standard graded $\kappa$-algebra is generated by $\mathfrak m/\mathfrak m^2$. The <Hilbert-Serre theorem> makes the cumulative Hilbert function eventually polynomial, which establishes the definition. For an <Artinian local ring>, the polynomial is constant and nonzero, so $d(R)=0$.
For every <prime chain> of length $s$, the <prime-chain lower bound for local length> proved in Question 4 gives $H_R(t)\geq c t^s$. A polynomial of degree $d(R)$ cannot satisfy this when $s>d(R)$. Thus \b[the requested inequality is]
$$
\boxed{\dim R\leq d(R).}
$$
For clarity, the <embedding dimension> is the different invariant
$$
e(R)=\dim_\kappa(\mathfrak m/\mathfrak m^2)
=\text{the minimal number of generators of }\mathfrak m.
$$
The equality follows from the <Nakayama lemma>. Applying Question 4 to $I=\mathfrak m$ gives $\dim R\leq e(R)$, while the polynomial-algebra surjection below also gives $d(R)\leq e(R)$. In particular, growth dimension and embedding dimension should be kept distinct.
A <regular local ring> is a <Noetherian local ring> with $e(R)=\dim R$. Let this common value be $e$ and choose a minimal generating set $x_1,\ldots,x_e$ of $\mathfrak m$. The initial forms give a surjective graded <ring homomorphism>
$$
\kappa[T_1,\ldots,T_e]\twoheadrightarrow\operatorname{gr}_{\mathfrak m}R.
$$
For $e=0$, the <Nakayama lemma> gives $\mathfrak m=0$, so $R$ is a <field>. Assume $e>0$. If the kernel contained a nonzero <homogeneous polynomial> $F$ of degree $q>0$, the cumulative Hilbert function of the target would be bounded by that of $\kappa[T_1,\ldots,T_e]/(F)$. Multiplication by $F$ is injective in the polynomial <integral domain>, so this bound is
$$
\binom{t+e-1}{e}-\binom{t-q+e-1}{e}=O(t^{e-1})\qquad(t>q).
$$
But $\dim R=e$ provides a <prime chain> of length $e$, and Question 4 gives $H_R(t)\geq ct^e$. This is a contradiction. Hence the <associated graded ring of a regular local ring> is
$$
\boxed{\operatorname{gr}_{\mathfrak m}R\cong\kappa[T_1,\ldots,T_e],}
$$
an <integral domain>.
Finally the <Krull intersection theorem> gives $\bigcap_j\mathfrak m^j=0$. One can see the needed separatedness directly: for the finitely generated <ideal> $K=\bigcap_j\mathfrak m^j$, the <Artin-Rees lemma> gives $K=\mathfrak mK$, and the <Nakayama lemma> gives $K=0$. Therefore each nonzero $a\in R$ has finite $\mathfrak m$-adic order $r$, with nonzero initial form in $\mathfrak m^r/\mathfrak m^{r+1}$. For nonzero $a,b$, their initial forms have nonzero product in the graded <integral domain>. It follows that $ab\ne0$, and indeed its order is the sum of their orders. \b[Thus $R$ is an <integral domain>.]
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