Let $(L,\leq)$ be a locally finite lattice, that is a partially ordered set such that any two pairs of elements $(x,y)$ have a least upper bound $x\vee y$, a greatest lower bound $x \wedge y$ and such that $\{z \in L : x\leq z\leq y\}$ is finite. We denote by $\mathcal{A}(L)$ its incidence algebra, that is the set of function $V : L^{2} \to \mathbb{R}$ such that $V(x,y)=0$ if $x\not \leq y$, the multiplication on this algebra being the convolution product
$$
V*W(x,z)=\sum_{x \leq y\leq z}V(x,y)W(y,z).
$$
This algebra admits a unit, the identity function $I(x,y)=1(x=y)$.
Remark. A function $V \in \mathcal{A}(L)$ can be equally viewed as a matrix $(V(x,y))_{x,y\in L}$. In that case, the convolution product $*$ is simply the matrix multiplication and identity is the identity matrix.
Examples. $(\mathbb{N}, \leq)$ is a lattice with $x\wedge y = \min (x,y)$ and $x \vee y=\max (x,y)$. Its incidence algebra $\mathcal{A}(\mathbb{N})$ can be identified with an "infinite" upper triangular matrix. In particular, $V$ is invertible if and only if every coefficient on the diagonal is non-zero.
$(\mathbb{N}^{*}, |)$ with the divisibility order is also a lattice with $x\wedge y = \operatorname{gcd}(x,y)$ and $x \vee y=\operatorname{lcm}(x,y)$. Its incidence algebra $\mathcal{A}(\mathbb{N})$ can be identified with an "infinite" upper triangular matrix whose coefficents are zero whenever $i \not| j$. Same as before, $V$ is invertible if and only if every coefficient on the diagonal is non-zero.
The latter result can be easily extended to more general lattice and we will take for granted that:
Existence of inverse. $V \in \mathcal{A}(L)$ is invertible if and only of $V(x,x)\neq 0$ for all $x\in L$.
Möbius function. As a consequence, $\zeta(x,y)=1(x\leq y)$ is invertible and its inverse (the Möbius function) is denoted by $\mu$.
If we let $L^{*}=\{f : L \to \mathbb{R}\}$, we see that $\mathcal{A}(L)$ acts on $L^{*}$ via the group action
$$
f.V(y)= \sum_{x \leq y}f(x)V(x,y), \quad f\in L^{*}, V \in \mathcal{A}(L)
$$
assuming $\{x \in L : x \leq y\}$ is finite for all $y\in L$.
In particular, one has
$$
f.\zeta(y)= \sum_{x \leq y}f(x), \quad f\in L^{*}, V \in \mathcal{A}(L).
$$
Möbius inversion formula. Let $f, g \in L^{*}$, then
$$
g(y)=\sum_{x \leq y}f(x) \quad \iff \quad f(y)=\sum_{x\leq y} g(x)\mu(x,y).
$$
Proof:
This is equivalent to $g=f.\zeta$ if and only if $f=g.\mu$, which is obviously true as $\zeta$ and $\mu$ are inverse.
Example: The Möbius inversion formula we obtained is very general and applies to many different settings. In $(\mathbb{N}^{*}, |)$ this is simply the well-known Möbius inversion formula in number theory. But you can also recover the principle of inclusion-exclusion (and many more formulas ...) by considering other lattices.
Back to the problem. Let $T$ be the linear operator defined by
$$
T : \left\{
\begin{aligned}
&L^{*} \to L^{*}\\
&f \to f.\zeta
\end{aligned}
\right.
$$
Reformulating the problem in our lattice framework, we need to prove that:
Proposition.
If $k\geq 1$ then $\operatorname{Ker}(T^{k}-I)=\{0\}$ on $(\mathbb{N}^{*},|)$.
The proof is divided into two steps.
Step 1.
For any lattice $(L,\leq)$,
$$
\operatorname{Ker}(T^{k}-I) = \operatorname{Ker}(T-I).
$$
Proof:
By the primary decomposition theorem,
$$
\operatorname{Ker}(T^{k}-I) = \operatorname{Ker}(T-I)\oplus \operatorname{Ker}(I+T+\dots+T^{k-1}).
$$
It remains to prove that $W=I+T+\dots+T^{k-1}$ is invertible. Note that
$$
W(f)=f.\underbrace{\left(\sum_{j=0}^{k-1}\zeta^{*_{j}}\right)}_{\Delta}
$$
where $\zeta^{*_{j}} = \zeta*\dots*\zeta$ $j$-times. By definition, for any $x\in L$, one has $\zeta^{*_{j}}(x,x) = \zeta^{*_{j-1}}(x,x)=\dots = 1$. Hence, $\Delta(x,x)=k \neq 0$. This implies that $V$ is invertible in $\mathcal{A}(L)$. In particular, $W$ is invertible and
$$
W^{-1}(f)=f.\Delta^{-1}.
$$
So, we just need to prove that $\operatorname{Ker}(T-I)=\{0\}$. Unfortunately, this is not always true. In fact, if $L$ is a finite lattice, then the latter is always false (why?) and we can also build infinite lattice for which this result does not hold either. This is where the particular structure of $(\mathbb{N}^{*},|)$ intervenes.
Step 2. $\operatorname{Ker}(T-I) = \{0\}$ on $(\mathbb{N}^{*},|)$.
Proof:
Let $f : \mathbb{N}^{*}\to \mathbb{R}$ such that $Tf=T$. We will prove that $f=0$. The preceding relationship is simply
$$
f(n)=\sum_{d|n} f(d).
$$
In particular, when $p$ is prime and $n\geq 1$, we have
$$
S_{p}:=\sum_{k=0}^{n-1} f(p^{k})=0.
$$
so that $f(p^{n})=S_{p+1}-S_{p}=0$.
Now the idea is to propagate this property to composite numbers. This is done inductively. Imagine you want to prove that $f(n)=0$ where $n=p^{a}q^{b}$, $p,q$ primes and $a,b\geq 0$. Assume you have proved that $f(n)=0$ for any such $n$ with $a+b\leq m$. Consider $n=p^{a}q^{b}$ with $a+b\leq m+1$. With $Tf(pn)=f(pn)$, we have
$$
\begin{aligned}
0=&\sum_{\substack{d | p^{a+1}q^{b},\\ d \neq p^{a+1}q^{b}}} f(d)\\
=&\sum_{d | p^{a+1}q^{b-1}}f(d)+\sum_{d | p^{a}q^{b}}f(d)-\sum_{d | p^{a}q^{b-1}}f(d)\\
=&f(p^{a+1}q^{b-1})+f(n)-f(p^{a}q^{q-1})\\
(a+b-1\leq m)=&f(n)+f(p^{a+1}q^{b-1}).
\end{aligned}
$$
Thus $f(n)=-f(p^{a+1}q^{b-1})$ and inductively, one has $f(n)=f(p^{a+2}q^{b-2})=\dots= (-1)^{b}f(p^{a+b})=0$.
Inducting on the number of primes in the decomposition of $n$, we can prove more generally that $f(n)=0$ for all $n\geq 1$. This finishes the proof.
To conclude, a question for which I don't have any answer: Characterize all lattices $(L, \leq)$ for which $\operatorname{Ker}(T-I)=\{0\}$.
