[proofplan] We expand the Dirichlet convolution with the constant one function, so the desired identity reduces to evaluating the divisor sum $\sum_{d \mid n}\mu(d)$. The case $n=1$ is immediate from the definition of $\mu(1)$. For $n \geq 2$, only squarefree divisors contribute to the sum; these are precisely the products of subsets of the distinct prime divisors of $n$. Their Möbius values form the finite alternating subset sum $\sum_{A}(-1)^{|A|}$, which factors as $(1-1)^r$ and hence vanishes. [/proofplan]

[step:Reduce the convolution identity to a divisor sum] Define the arithmetic function $S: \mathbb{N} \to \mathbb{C}$ by \begin{align*} S(n) = (\mu * \mathbf{1})(n). \end{align*} By the definition of Dirichlet convolution and by $\mathbf{1}(m)=1$ for every $m \in \mathbb{N}$, \begin{align*} S(n) &= \sum_{d \mid n} \mu(d)\mathbf{1}(n/d)\\ &= \sum_{d \mid n} \mu(d). \end{align*} Thus it suffices to prove that this divisor sum equals $\varepsilon(n)$ for every $n \in \mathbb{N}$. [/step]

[step:Evaluate the divisor sum at $n=1$] The positive divisors of $1$ consist only of $1$. Since the Möbius function is defined by $\mu(1)=1$, we have \begin{align*} S(1) = \sum_{d \mid 1}\mu(d) = \mu(1)=1=\varepsilon(1). \end{align*} [/step]

[step:Identify the squarefree divisors that can contribute when $n \geq 2$]Fix $n \in \mathbb{N}$ with $n \geq 2$. Let $p_1,\dots,p_r$ be the distinct prime divisors of $n$, where $r \geq 1$. For each subset $A \subset \{1,\dots,r\}$, define the divisor $d_A \in \mathbb{N}$ by \begin{align*} d_A = \prod_{i \in A} p_i, \end{align*} with the empty product interpreted as $d_{\varnothing}=1$. If $d \mid n$ and $\mu(d) \neq 0$, then $d$ is squarefree, so every prime divisor of $d$ occurs to exponent $1$. Since every prime divisor of $d$ is one of the primes $p_1,\dots,p_r$, there is a unique subset $A \subset \{1,\dots,r\}$ such that $d=d_A$. Conversely, each $d_A$ is squarefree and divides $n$, hence \begin{align*} \mu(d_A)=(-1)^{|A|}. \end{align*} Therefore \begin{align*} \sum_{d \mid n}\mu(d) = \sum_{A \subset \{1,\dots,r\}}(-1)^{|A|}. \end{align*}[/step]

[guided]The Möbius function vanishes exactly on integers divisible by the square of a prime. Therefore, in the divisor sum \begin{align*} \sum_{d \mid n}\mu(d), \end{align*} we only need to keep track of the squarefree divisors of $n$. Let $p_1,\dots,p_r$ be the distinct prime divisors of $n$. Since $n \geq 2$, at least one prime divides $n$, so $r \geq 1$. For each subset $A \subset \{1,\dots,r\}$, define \begin{align*} d_A = \prod_{i \in A} p_i, \end{align*} where the empty product is $d_{\varnothing}=1$. This construction gives a squarefree divisor of $n$: each prime $p_i$ with $i \in A$ appears once, and no other prime appears. Now let $d \mid n$ satisfy $\mu(d)\neq 0$. By the definition of the Möbius function, $d$ is squarefree. Because $d$ divides $n$, every prime divisor of $d$ must be among $p_1,\dots,p_r$. Since $d$ is squarefree, it is exactly the product of the distinct primes among $p_1,\dots,p_r$ that divide it. Thus there is a unique subset $A \subset \{1,\dots,r\}$ such that $d=d_A$. For such a divisor $d_A$, the Möbius function counts the parity of the number of prime factors: \begin{align*} \mu(d_A)=(-1)^{|A|}. \end{align*} All other divisors have Möbius value $0$. Hence the whole divisor sum reduces to the finite subset sum \begin{align*} \sum_{d \mid n}\mu(d) = \sum_{A \subset \{1,\dots,r\}}(-1)^{|A|}. \end{align*}[/guided]

[step:Evaluate the alternating subset sum] The finite product expansion gives \begin{align*} \prod_{i=1}^{r}(1-1) &= \sum_{A \subset \{1,\dots,r\}} \prod_{i \in A}(-1)\\ &= \sum_{A \subset \{1,\dots,r\}}(-1)^{|A|}. \end{align*} Since $r \geq 1$, the left-hand side is $0$. Therefore \begin{align*} \sum_{d \mid n}\mu(d)=0=\varepsilon(n). \end{align*} Combining this with the case $n=1$, we have $S(n)=\varepsilon(n)$ for every $n \in \mathbb{N}$. Since $S=\mu * \mathbf{1}$, this proves \begin{align*} \mu * \mathbf{1}=\varepsilon. \end{align*} [/step]