Hilbert-Samuel multiplicity as the factorial-scaled leading coefficient

Let $(R,\mathfrak{m})$ be a Noetherian local ring, let $M$ be a finite $R$-module, and let $I\subseteq \mathfrak{m}$ be an ideal of definition for $M$.

If $M=0$, define $e_I(M):=0.$

If $M\mathrm{\ne }0$, suppose the Hilbert-Samuel function ${\chi }_{I,M}(n)$ agrees for all sufficiently large $n$ with a polynomial $P_{I,M}\in \mathbb{Q}[n]$. Such a polynomial is unique. Because $I\subseteq \mathfrak{m}$ and $M\mathrm{\ne }0$, the quotient $M\mathrm{/}{I}^{n+1}M$ is nonzero for every $n$, so $P_{I,M}$ is not the zero polynomial. Let $d=\mathrm{deg}{P}_{I,M}$.

The Hilbert-Samuel multiplicity of $M$ with respect to $I$ is $e_I(M):=d!\cdot (\text{leading coefficient of }{P}_{I,M})\mathrm{.}$

Equivalently, when $M\mathrm{\ne }0$ and $P_{I,M}(n)=\frac{e_I(M)}{d!}{n}^d+\text{lower-degree terms},$ then $e_I(M)$ is the scaling of the top term. It is an integer: the $d$th forward difference of the integer-valued polynomial $P_{I,M}$ at any sufficiently large integer is $d!$ times its leading coefficient.

The quoted Nakayama step is choice-free here. If $M\mathrm{/}{I}^{n+1}M=0$, then $I^{n+1}M=M$. By Determinant trick for Nakayama, some $a\in I^{n+1}\subseteq \mathfrak{m}$ satisfies $(1-a)M=0$. In a local ring $1-a$ is a unit, forcing $M=0$, contrary to the hypothesis.