Length of regular local ring Let $A$ be a regular local ring of dimension $n$ with maximal ideal $\mathcal m$.
Then one can consider the $A-$module $A/\mathcal m^i$ for all natural numbers $i$.
My question is simply what length this module has. For $i=0$ it is clearly of length $1$.
But what about $i=2$ and bigger? Can one write down a formula for the length depending just on $n$?
And do you know a reference for this problem?
 A: $\newcommand\m{\mathfrak{m}}$
For regular local rings $A$ of dimension $n$ one has the formula:
$$\mathrm{length}(A/\m^{i+1}) = \binom{i+n}{n}.$$
More generally,
let $(A,\m)$ be any Noetherian local ring.
Associated to $A$ is the graded $k = A/\m$-algebra 
$$\mathrm{gr}(A) =
\bigoplus_{j=0}^{\infty} \mathrm{gr}^j(A),$$
where $\mathrm{gr}^j(A):=\m^j/\m^{j+1}$. It is a fact that $\mathrm{gr}(A)$
is a local $k$-algebra of dimension equal to $n = \dim(A)$.
The length of $A/\m^{i+1}$ is equal to the sums of the lengths of
$\m^j/\m^{j+1}$ for $j \le i$. Since these latter modules are vector spaces,
this is the same as saying that:
$$\chi_A(i) := \mathrm{length}(A/\m^{i+1}) =
\sum_{j=0}^{i} \dim_k \mathrm{gr}^j(A).$$
The main fact to know is that  $\chi_A(i)$
is a polynomial for sufficiently large $i$. This polynomial is known
as the Hilbert-Samuel polynomial; it has degree $n$ and leading coefficient $1/n!$.
If $A$ is regular, however, then $\dim \m/\m^2 = n$, and so by Nakayama's
lemma, $\mathrm{gr}(A)$ is a quotient of $k[x_1,x_2, \ldots, x_n]$.
Since any non-trivial quotient of this has dimension less than $n$, it follows that
$$\mathrm{gr}(A) \simeq k[x_1,x_2, \ldots, x_n].$$
From this it follows that
the Hilbert-Samuel polnomial is exactly:
$$\binom{i+n}{n},$$
and moreover that $\chi_A(i)$ is given by this formula for all $i \ge 0$.
