 Solution

Let $Q=Q_{n+1}$. By polynomial division, every $q\in {\mathcal{P}}_{n+1+k}$ has a unique decompositionEvery node is a zero of $Q$, so $I_n(Qs)=0$. The assumed degree-$n$ exactness also gives $I_n(r)=I(r)$. HenceTherefore $I_n(q)=I(q)$ for every $q\in {\mathcal{P}}_{n+1+k}$ if and only iffor every $s\in {\mathcal{P}}_k$. This proves the nodal-polynomial criterion for quadrature exactness.