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Here I gave a proof for $\sum_{k=0}^n\binom nk(-1)^k=0$ based on the fact that $\mu*1=\varepsilon$ (the Dirichlet convolution identity). I am wondering if using a similar technique we can prove that $\sum_{k=0}^n\binom nka^k=(1+a)^n$ (for all $a\in\mathbb Z$, say) and thus deduce the binomial theorem.

I'd be happy if we have this for at least $n+1$ values for $a$ (because then it follows for all $a$).

There could be a modified Möbius-function with a similar property as $\sum_{d\mid n}\mu(d)=0$, which somehow gives rise to the more general $\sum_{k=0}^n\binom nka^k=(1+a)^n$...

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Bart Michels
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Here's a proof using a construction similar to Dirichlet convolution:

Fact. Let $D_n$ denote the set of (positive) divisors of $n$. Let $f$ be a multiplicative function with values in $\mathbb N^+$ that sends prime powers to prime powers and that preserves the coprimality relation. Let $g$ be any multiplicative function. Suppose $f^{-1}(D_n)$ is finite for all $n$. Then $$n\mapsto\sum_{d\in f^{-1}(D_n)}g(d)$$ is multiplicative.
(In fact it suffices to require that $\sum_{d\in f^{-1}(D_n)}g(d)$ converges absolutely for all $n$.) Note that for $f(n)=n$ we get the usual $g*1$.

Proof of the binomial theorem. Let $x\in\mathbb C$. $\newcommand\rad{\operatorname{rad}}$Let $f=\rad$ (the radical, the product of prime factors) and define a multiplicative $g$ on prime powers by $g(p)=x$ and $g(p^k)=0$ for $k>1$. Let $N$ be the product of $n$ distinct prime numbers. By the combinatorial definition of binomial coefficients, $$\sum_{d\in\rad^{-1}(D_N)}g(d)=\sum_{k=0}^n\binom nkx^k.$$

On the other hand, using multiplicativity it equals $$\left(\sum_{d\in\rad^{-1}(2017)}g(d)\right)^n=(1+x)^n.$$

Proof of the fact. Let $S_n=f^{-1}(D_n)$. By multiplicativity of $g$ (and some issues with convergence, if you wish) it suffices that $S_nS_m=S_{nm}$ and $|S_n||S_m|=|S_{nm}|$ for coprime $n,m$. We have $\subset$ because coprime numbers have coprime preimages, $\supset$ because $f$ sends coprime prime powers to coprime prime powers, and $\cdot:S_n\times S_m\to S_{nm}$ is injective again because the elements of $S_n$ are coprime with those of $S_m$.

Bart Michels
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