Binomial inversion ($b_n={\sum }_{k=0}^n(-1{)}^{n-k}(\genfrac{}{}{0ex}{}{n}{k})a_k⟺a_n={\sum }_{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})b_k$)

These two finite-sum transformations are inverses over any commutative ring. Substitute one sum into the other and use $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left(\genfrac{}{}{0ex}{}{k}{j}\right)=\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{k-j}\right)$. The inner alternating sum is $(1-1{)}^{n-j}$ by the binomial theorem, so only $j=n$ survives. In a Mahler expansion, $b_n$ is the nth finite difference of the sampled values $a_k$.