Product of Taylor polynomials 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. 

You're trying to do things the hard way.  To save myself some typing, I'll write $ a+h = x$,  $P^k_{f,a}(a+h) = T_f(x)$ and similarly for $g$ and $fg$.  We
have $f(x) = T_f(x) + o(|h|^k)$ and similarly for $g$ and $fg$.
So $$T_f(x) T_g(x) - T_{fg}(x) = (f(x) + o(|h|^k))(g(x) + o(|h|^k)) - (f(x) g(x) + o(|h|^k)) = o(|h|^k)$$
