Tensor-product Gaussian quadrature preserves coordinatewise exactness

For a coordinate monomial $P_{\alpha }(x)={\prod }_{j=1}^d{x}_j^{{\alpha }_j}$, substitute the rule definition [F1] and distribute the finite sums: $Q_{\otimes }(P_{\alpha })={\prod }_{j=1}^d({\sum }_{i_j=1}^{m_j}{w}_{i_j}^{(j)}(x_{i_j}^{(j)}{)}^{{\alpha }_j})\mathrm{.}$

Since $0\le {\alpha }_j\le 2m_j-1$, [F3] makes the $j$th factor in step 1.1 equal to $M_{{\alpha }_j}^{(j)}$. Their product is exactly $I_{\otimes }(P_{\alpha })$ by [F2].

A coordinatewise tensor polynomial is a finite linear combination of these monomials. Both $Q_{\otimes }$ and $I_{\otimes }$ are linear finite sums, so step 2.1 extends term by term and proves $Q_{\otimes }(P)=I_{\otimes }(P)$.

Each one-dimensional node list has $m_j$ distinct entries by The Gauss orthogonal polynomial has simple roots inside the interval. Two different tuples differ in at least one coordinate, whose node entries are distinct, so they give different points. The nested index loops in [F1] enumerate $m_1$ tuples for $d=1$; appending coordinate $d+1$ replaces each prior tuple by exactly $m_{d+1}$ distinct extensions. Iterating this finite recurrence gives exactly ${\prod }_j{m}_j$ nodes, hence $m^d$ when every $m_j=m\ge 2$.