On the Hilbert function of a numerical semigroup Recall that a numerical semigroup $S$ is a submonoid of the non-negative integers $\mathbb Z_{\geq 0}$ whose relative complement $\mathbb Z_{\geq 0} \setminus S$ is finite. Observe that the collection $S^*$ of nonzero elements of $S$ constitutes an ideal of $S$ in the sense that $$S^* \supseteq \{s + t \mid s \in S^* \text{ and } t \in S\} = S^* + S;$$ because of this, we refer to $S^*$ as the maximal ideal of $S.$
Generally, if $n$ is a non-negative integer, then the $n$-fold sum of $S^*$ is given by $nS^* = \{s_1 + \cdots + s_n \mid s_1, \dots, s_n \in S^*\}.$ One can verify that for each positive integer $n,$ the sets $nS^* \setminus (n + 1) S^*$ are finite; their cardinalities give rise to the Hilbert function of $S,$ defined by $H_S(n) = \#(nS^* \setminus (n + 1) S^*).$
Given an infinite field $k,$ we may also define the numerical semigroup ring $k [\![S]\!]$ corresponding to $S.$ It is well-known that the Hilbert-Samuel multiplicity of $k [\![S]\!]$ is exactly $\min(S^*)$; the latter is therefore called the multiplicity of $S,$ denoted by $\operatorname e(S).$ Bearing this in mind, by definition of the multiplicity of $k [\![S]\!],$ it follows that $H_S(n) = \operatorname e(S)$ for all integers $n \gg 0.$ If I recall correctly, this holds for all $n \geq \operatorname e(S).$

Question. Is there a way to see that $\#(nS^* \setminus (n + 1) S^*)$ eventually stabilizes without resorting to computations involving the numerical semigroup ring of $S$? Put another way, is there a purely numerical semigroup-theoretic proof that $\#(nS^* \setminus (n + 1) S^*) = \operatorname e(S)$ for all integers $n \geq \operatorname e(S)$?

I have noticed that this is tacitly acknowledged throughout the literature, but as yet, I have not seen it formally addressed. Using the numericalsgps package of the GAP System, I have accrued a substantial amount of data that suggests this could even be improved to show that $\#(nS^* \setminus (n + 1) S^*) = \operatorname e(S)$ for all integers $n \geq \operatorname e(S) - 1,$ but it is not obvious to me that this holds in general. I would appreciate any comments, suggestions, or references. Thanks in advance for your time and consideration.
 A: I believe there is an easier proof.
Proof. By induction, it suffices to prove the case that $eS^* = e + (e - 1) S^*$: indeed, if we assume that $(n + 1) S^* = e + nS^*$ for some integer $n \geq e,$ then it holds that $$(n + 2) S^* = (n + 1) S^* + S^* = e + nS^* + S^* = e + (n + 1) S*.$$
Because the containment $eS^* \supseteq e + (e - 1) S^*$ is clear, we will prove the other containment. Equivalently, we will show that for every element $x \in eS^*,$ we have that $x - e$ lies in $(e - 1) S^*.$ Observe that every element $x \in e S^*$ can be written as a sum $x = s_1 + \cdots + s_e$ of $e$ positive integers $s_1, \dots, s_e.$ By the Pigeonhole Principle, we can write $x = \sum_{s \in A} s + \sum_{s \notin A} s$ for some nonempty subset $A \subseteq \{s_1, \dots, s_e\}$ such that $\sum_{s \in A} s$ is divisible by $e$: among the $e$ sums $s_1, s_1 + s_2, \dots, s_1 + \cdots + s_e,$ either one of them is divisible by $e$ or some two of them must have the same remainder modulo $e,$ in which case their difference is divisible by $e.$ Ultimately, we conclude that $x - e = {\left(\sum_{s \in A} s - e\right)} + \sum_{s \notin A} s$ lies in $(e - 1) S^*,$ as desired. QED.
A: Let $\text{e}$ be the minimum of $S^*$. Call a height function a function from $\mathbb Z/\text{e} \mathbb Z$ to the natural numbers.
The height function $A$ for the numerical semigroup $S^*$ is defined as follows: $A(x)$ is the minimum element in $S^*$ that is in the congruence class of $x$.
The point is that the height function together with $\text{e}$ completely determines the numerical semigroup, so the problem reduces to studying the height functions of $nS^*$ and $(n+1)S^*$. Let $A_n$ be the height function of $nS^*$ subtracted by $\text{e}(n-1)$. Then $A_1$ is the height function of $S^*$. For every $n\geq 1$, $A_n(0)=\text{e}$, and $A_n(x) > \text{e}$ for any nonzero $x$.
Lemma. The height function $A_{n+1}$ is determined from $A_n$ by the formula $A_{n+1}(x)=\underset{y+z=x} \min (A_n(y)+A_1(z)-\text{e})$. (Note that $y$ and $z$ are elements of $\mathbb Z/\text{e} \mathbb Z$.)
Corollary. $A_{n+1}(x) \leq A_n(x)$ for every $x$. This is because we can take $y=x$ and $z=0$. Since $A_{n}(x)$ is positive and nonincreasing w.r.t $n$ for every $x$, $A_n$ stabilizes for sufficiently large $n$. So we have $\#(nS^∗ \backslash(n+1)S^∗)=\text{e}$ for sufficiently large $n$.
Repeating the height function iteration gives $A_n(x)=\underset{x_1+x_2+\cdots+x_n=x}{\min} A_1(x_1)+A_1(x_2)+\cdots+A_1(x_n)-\text{e}(n-1)$.  (Note that $x_1,\cdots,x_n$ are elements of $\mathbb Z/\text{e} \mathbb Z$.)
Now I will prove $A_\text{e}=A_{\text{e}-1}$. By the corollary, only $A_{\text{e}-1} \leq A_\text{e}$ needs to be proven.
Let $x$ be any element in $\mathbb Z/\text{e} \mathbb Z$ and $A_\text{e}(x)$ be minimized at $x_1, \cdots, x_\text{e}$ where $x_1, \cdots, x_\text{e} \in \mathbb Z/\text{e} \mathbb Z$ sums to $x$. Consider the sequence $x_1,x_2, \cdots, x_\text{e}, 0, 0, \cdots 0$, i.e. appending $\text{e}-1$ zeros after $x_1,x_2, \cdots, x_\text{e}$. By the Erdős–Ginzburg–Ziv theorem, there exists a length-$\text{e}$ subsequence of this sequence that sums to $0$. Removing the appending zeros, it follows that a nonempty subsequence $x_{m_1}, \cdots, x_{m_l}$ of $x_1, \cdots, x_\text{e}$ sums to $0$. Since $x_1, \cdots, x_\text{e}$ minimizes $A_\text{e}(x)$, it follows that $x_{m_1}, \cdots, x_{m_l}$ are all $0$, since otherwise setting all of them to $0$ gives a smaller value of $A_\text{e}(x)$, which contradicts the minimality of $A_\text{e}(x)$.
But now we have $A_{\text{e}-l}(x) = A_1(y_1)+A_1(y_2)+ \cdots A_1(y_{\text{e}-l}) - \text{e} (\text{e}-l-1)$ where $y_1, \cdots, y_{\text{e}-l}$ is the complementary sequence of $x_{m_1}, \cdots, x_{m_l}$ in $x_1, \cdots, x_\text{e}$. This is because RHS $=$ $A_{\text{e}}(x) \leq$ LHS $\leq$ RHS, where the $=$ is because $x_{m_1}= \cdots =x_{m_l}=0$, the first $\leq$ is by the corollary, and the second $\leq $ is by the definition of $A_{\text{e}-l}(x)$. Thus $A_{\text{e}-l}(x)=A_{\text{e}}(x)$, and by the corollary, $A_{\text{e}-1}(x)=A_{\text{e}}(x)$.
This shows that $(n+1)S^∗$ is a shift of $nS^∗$ by $\text{e}$ for all integers $n≥\text{e}-1$, so $\#(nS^∗ \backslash(n+1)S^∗)=\text{e}$ for $n≥\text{e}-1$.
