An arithmetic function has a Dirichlet inverse exactly when its value at $1$ is nonzero

Suppose $g$ satisfies $f\ast g=\epsilon$. Evaluating at $n=1$ gives $f(1)g(1)=\epsilon (1)=1$, so $f(1)\mathrm{\ne }0$ and $g(1)=1\mathrm{/}f(1)$.

Now assume $f(1)\mathrm{\ne }0$. Since $\mathbb{C}$ is a field by $\mathbb{C}=\mathbb{R}[x]\mathrm{/}(x^2+1)$ is a field, every element is uniquely $a+bi$, and every nonzero element has inverse $(a-bi)\mathrm{/}(a^2+b^2)$, set $g(1):=1\mathrm{/}f(1)$. Then $(f\ast g)(1)=f(1)g(1)=1=\epsilon (1)$.

Fix $n>1$ and assume inductively that $g(d)$ has already been defined for every $d<n$ so that $(f\ast g)(m)=\epsilon (m)$ for every $m<n$. Every proper positive divisor $d$ of $n$ satisfies $d<n$ by If $d\mid a$ and $a\mathrm{\ne }0$ then $d\mathrm{\ne }0$ and $\mathrm{\mid }d\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }$; hence the set of divisors of a nonzero integer is bounded above by $\mathrm{\mid }a\mathrm{\mid }$, so in the identity ${\sum }_{d\mid n}f(d)g(n\mathrm{/}d)=\epsilon (n)=0$ the only unknown term is $f(1)g(n)$. Solving for that term and reindexing the proper divisors by $d\leftrightarrow n\mathrm{/}d$ gives the displayed recursion, and with that choice the equality $(f\ast g)(n)=0$ holds.

Apply Strong (complete) induction to the property on natural numbers $r$ that the displayed recursion uniquely defines $g(m)$ and makes $(f\ast g)(m)=\epsilon (m)$ for every $1\le m\le r+1$. Step 1.2 gives the base case $r=0$, and step 1.3 gives the strong induction step. Therefore there is a unique arithmetic function $g$ with $f\ast g=\epsilon$ for every positive integer $n$.

By Arithmetic functions form a commutative ring under pointwise addition and Dirichlet convolution, Dirichlet convolution is commutative, so $g\ast f=f\ast g=\epsilon$. Together with step 1.1, this proves the equivalence, the recursive formula, and uniqueness.