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 $\sum_{k=0}^n\binom nka^k=(1+a)^n$...