Another way to demonstrate binomial inversion

A clever method to collect evidence
there $f(i)=\sum_{j=i}^n\binom{j}{i}g(j)$
Certificate $g(i)=\sum_{j=i}^n(-1)^{j-i}\binom{j}{i}f(i)$
Consider $f,g$ generation function
$G (x) = \ sum_ {i = 0} ^ (i) x ^ i$
then $G(x+1)=\sum_{i=0}^ng(i)\sum_{j=0}^ix^j\binom{i}{j}=\sum_{j=0}^nx^j\sum_{i=j}^n\binom{i}{j}g(i)=F(x)$
so $G(x)=F(x-1)=\sum_{i=0}^nf(i)\sum_{j=0}^i(-1)^{j-i}\binom{i}{j}x^j=\sum_{i=0}^nx^i\sum_{j=i}^n\binom{j}{i}(-1)^{j-i}f(j)$