Let $f$ be an arithmetic function $f^{*k}:=f*...*f$ k times be the k-th Dirichlet convolution. An arithmetic function $g$ is said to be a k-th Dirichlet root of an arithmetic function $f$ if $g^{*k}=f$. Show that if $f(1)\neq 0$ then has exactly k Dirichlet roots. Moreover, if $f$ is multiplicative then exactly one of the k roots is also multiplicative.
I've thought about defining $g$ inductively but I get a mess of a condition when applying the definition of the Dirichlet convolution. Thanks in advanced for any help.
You have the right idea, you just need to use a nicer definition of the Dirichlet convolution: $$f*g(n)=\sum_{ab=n}f(a)g(b)$$ So $$f^{*k}(n)=\sum_{a_1\cdots a_k=n}f(a_1)\cdots f(a_k)$$ Then the induction is pretty straightforward: there are $k$ choices for $g(1)$, and the recurrence gives us a linear equation for $g(n)$ in terms of the previously selected values.