I'm trying to prove the following proposition:
Let $U\in R^n$ be open, and $f,g\colon U\to R$ be $C^k$ functions, then the Taylor polynomial of $fg$ is computed as $P_{f,a}^k(a+\vec{h})\cdot P_{g,a}^k(a+\vec{h})$ and discarding the terms of degree > $k$, where $P_{f,a}^k(a+\vec{h})$ denotes the degree $k$ Taylor polynomial of $f$ at $a$.
And here's what I've got so far:
$P_{fg,a}^k=\sum_{m=0}^k\sum_{I\in I_n^m}\frac{1}{I!}D_I(fg)(a)$ using the definition of Taylor polynomial of multi-variable functions. ($I=(i_1,i_2,...,i_n)$ and $D_If=D_1^{i_1}D_2^{i_2}...D_n^{i_n}f)$.
Then I think $D_I(fg)$ can be written as $D_I(f)g+fD_I(g)$ using the product rule so that $P_{fg,a}^k(a+\vec{h})=P_{f,a}^k(a+\vec{h})\cdot g+f\cdot P_{g,a}^k(a+\vec{h})$
but I can't figure out how this leads to the final result.