[proofplan] We recast the divisor-sum hypothesis as an identity for Dirichlet convolution. The central algebraic fact is that the Möbius function is the convolution inverse of the constant-one arithmetic function. After proving this inverse identity directly from the definition of $\mu$, associativity of the finite divisor sums gives $\mu * g = f$, and evaluating that identity at $n$ gives the inversion formula. [/proofplan] [step:Define Dirichlet convolution and rewrite the hypothesis] Let $\mathbf{1}: \mathbb{N} \to \mathbb{C}$ denote the constant arithmetic function given by $\mathbf{1}(n)=1$ for every $n \in \mathbb{N}$. For arithmetic functions $a,b: \mathbb{N} \to \mathbb{C}$, define their Dirichlet convolution $a*b: \mathbb{N} \to \mathbb{C}$ by \begin{align*} (a*b)(n)=\sum_{d\mid n} a(d)b(n/d). \end{align*} Then the hypothesis says exactly that \begin{align*} g(n)=\sum_{d\mid n} f(d)\mathbf{1}(n/d)=(f*\mathbf{1})(n) \end{align*} for every $n \in \mathbb{N}$. Hence $g=f*\mathbf{1}$ as arithmetic functions. [guided] The point of introducing Dirichlet convolution is to turn the divisor-sum formula into an algebraic product. Define the arithmetic function \begin{align*} \mathbf{1}: \mathbb{N} &\to \mathbb{C} \\ n &\mapsto 1. \end{align*} For arithmetic functions $a,b: \mathbb{N} \to \mathbb{C}$, their Dirichlet convolution is the arithmetic function \begin{align*} a*b: \mathbb{N} &\to \mathbb{C} \\ n &\mapsto \sum_{d\mid n} a(d)b(n/d). \end{align*} The sum is finite because every positive integer has only finitely many positive divisors. Applying this definition with $a=f$ and $b=\mathbf{1}$ gives, for every $n \in \mathbb{N}$, \begin{align*} (f*\mathbf{1})(n) &=\sum_{d\mid n} f(d)\mathbf{1}(n/d) \\ &=\sum_{d\mid n} f(d). \end{align*} By the hypothesis, the final expression is $g(n)$. Thus $g(n)=(f*\mathbf{1})(n)$ for every $n \in \mathbb{N}$, which is the function identity $g=f*\mathbf{1}$. [/guided] [/step] [step:Prove that the Möbius function inverts the constant-one function] Let $\varepsilon: \mathbb{N} \to \mathbb{C}$ denote the identity arithmetic function for Dirichlet convolution, defined by \begin{align*} \varepsilon(n)= \begin{cases} 1, & n=1,\\ 0, & n>1. \end{cases} \end{align*} We claim that \begin{align*} \mu * \mathbf{1}=\varepsilon. \end{align*} Indeed, for $n=1$, \begin{align*} (\mu*\mathbf{1})(1)=\mu(1)\mathbf{1}(1)=1. \end{align*} If $n>1$, let $p_1,\dots,p_k$ be the distinct prime divisors of $n$, so $k\geq 1$. Since $\mu(d)=0$ unless $d$ is squarefree, only the squarefree divisors of $n$ contribute. These are exactly the products $\prod_{i\in S}p_i$ indexed by subsets $S\subseteq\{1,\dots,k\}$, and for such a divisor $\mu(\prod_{i\in S}p_i)=(-1)^{|S|}$. Therefore \begin{align*} (\mu*\mathbf{1})(n) &=\sum_{d\mid n}\mu(d) \\ &=\sum_{S\subseteq\{1,\dots,k\}}(-1)^{|S|} \\ &=(1-1)^k \\ &=0. \end{align*} Thus $(\mu*\mathbf{1})(n)=\varepsilon(n)$ for every $n \in \mathbb{N}$. [guided] We need the fact that $\mu$ cancels the constant-one function under Dirichlet convolution. Define \begin{align*} \varepsilon: \mathbb{N} &\to \mathbb{C} \\ n &\mapsto \begin{cases} 1, & n=1,\\ 0, & n>1. \end{cases} \end{align*} This is the identity arithmetic function for Dirichlet convolution. We now prove $\mu*\mathbf{1}=\varepsilon$ by evaluating both sides at an arbitrary $n \in \mathbb{N}$. If $n=1$, then the only divisor of $1$ is $1$, and the defining value $\mu(1)=1$ gives \begin{align*} (\mu*\mathbf{1})(1) &=\sum_{d\mid 1}\mu(d)\mathbf{1}(1/d) \\ &=\mu(1)\mathbf{1}(1) \\ &=1. \end{align*} This equals $\varepsilon(1)$. Now suppose $n>1$. Let $p_1,\dots,p_k$ be the distinct prime divisors of $n$. Since $n>1$, we have $k\geq 1$. By the definition of the Möbius function, $\mu(d)=0$ whenever $d$ is divisible by the square of a prime. Hence, in the divisor sum \begin{align*} (\mu*\mathbf{1})(n)=\sum_{d\mid n}\mu(d), \end{align*} only squarefree divisors $d$ contribute. The squarefree divisors of $n$ are precisely the products \begin{align*} \prod_{i\in S}p_i \end{align*} where $S$ ranges over all subsets of $\{1,\dots,k\}$. For such a divisor, the definition of $\mu$ gives \begin{align*} \mu\left(\prod_{i\in S}p_i\right)=(-1)^{|S|}. \end{align*} Therefore \begin{align*} (\mu*\mathbf{1})(n) &=\sum_{S\subseteq\{1,\dots,k\}}(-1)^{|S|} \\ &=\sum_{j=0}^{k}\binom{k}{j}(-1)^j \\ &=(1-1)^k \\ &=0. \end{align*} This equals $\varepsilon(n)$ for every $n>1$. Combining the cases $n=1$ and $n>1$ proves $\mu*\mathbf{1}=\varepsilon$. [/guided] [/step] [step:Use associativity to isolate $f$] We first record the associativity computation for Dirichlet convolution. For arithmetic functions $a,b,c: \mathbb{N}\to\mathbb{C}$ and $n\in\mathbb{N}$, \begin{align*} ((a*b)*c)(n) &=\sum_{e\mid n}\left(\sum_{d\mid e}a(d)b(e/d)\right)c(n/e) \\ &=\sum_{\substack{d,m,r\in\mathbb{N}\\ dmr=n}}a(d)b(m)c(r) \\ &=\sum_{d\mid n}a(d)\left(\sum_{m\mid n/d}b(m)c((n/d)/m)\right) \\ &=(a*(b*c))(n). \end{align*} All sums are finite divisor sums, so the rearrangement is valid. Using $g=f*\mathbf{1}$, associativity, and $\mu*\mathbf{1}=\varepsilon$, we obtain \begin{align*} \mu*g &=\mu*(f*\mathbf{1}) \\ &=(\mu*\mathbf{1})*f \\ &=\varepsilon*f. \end{align*} Finally, for every $n\in\mathbb{N}$, \begin{align*} (\varepsilon*f)(n) &=\sum_{d\mid n}\varepsilon(d)f(n/d) \\ &=\varepsilon(1)f(n) \\ &=f(n), \end{align*} because $\varepsilon(d)=0$ for every divisor $d>1$. Hence $\mu*g=f$. [guided] We now use the algebraic structure of Dirichlet convolution. First, we verify associativity directly because the proof depends on changing the grouping of a triple convolution. Let $a,b,c: \mathbb{N}\to\mathbb{C}$ be arithmetic functions. For each $n\in\mathbb{N}$, the definition of convolution gives \begin{align*} ((a*b)*c)(n) &=\sum_{e\mid n}(a*b)(e)c(n/e) \\ &=\sum_{e\mid n}\left(\sum_{d\mid e}a(d)b(e/d)\right)c(n/e). \end{align*} In the inner sum, write $m=e/d$ and write $r=n/e$. Then $d,m,r\in\mathbb{N}$ and $dmr=n$. Conversely, every factorization $dmr=n$ determines $e=dm$, with $e\mid n$ and $d\mid e$. Therefore \begin{align*} ((a*b)*c)(n) &=\sum_{\substack{d,m,r\in\mathbb{N}\\ dmr=n}}a(d)b(m)c(r). \end{align*} Since this is a finite sum over factorizations of $n$, we may regroup it by first choosing $d\mid n$ and then choosing $m\mid n/d$: \begin{align*} \sum_{\substack{d,m,r\in\mathbb{N}\\ dmr=n}}a(d)b(m)c(r) &=\sum_{d\mid n}a(d)\left(\sum_{m\mid n/d}b(m)c((n/d)/m)\right) \\ &=(a*(b*c))(n). \end{align*} Thus Dirichlet convolution is associative. Now we apply this associativity to the identity $g=f*\mathbf{1}$. Convolving both sides on the left by $\mu$ gives \begin{align*} \mu*g=\mu*(f*\mathbf{1}). \end{align*} By associativity and commutativity of the displayed finite divisor sums, we may group the right-hand side as \begin{align*} \mu*(f*\mathbf{1})=(\mu*\mathbf{1})*f. \end{align*} The previous step proved $\mu*\mathbf{1}=\varepsilon$, so \begin{align*} \mu*g=\varepsilon*f. \end{align*} It remains to check that $\varepsilon$ really acts as an identity. For every $n\in\mathbb{N}$, \begin{align*} (\varepsilon*f)(n) &=\sum_{d\mid n}\varepsilon(d)f(n/d). \end{align*} The only divisor $d$ with $\varepsilon(d)\neq 0$ is $d=1$, and $\varepsilon(1)=1$. Hence \begin{align*} (\varepsilon*f)(n)=f(n). \end{align*} Therefore $\mu*g=f$ as arithmetic functions. [/guided] [/step] [step:Evaluate the convolution identity to obtain the inversion formula] Since $\mu*g=f$, evaluating at an arbitrary $n\in\mathbb{N}$ gives \begin{align*} f(n) &=(\mu*g)(n) \\ &=\sum_{d\mid n}\mu(d)g(n/d). \end{align*} This is the desired Möbius inversion formula for every $n\in\mathbb{N}$. [/step]