TITLE: Proof of product rule $(fg)^{(n)}$
QUESTION [7 upvotes]: I went through induction proofs and they are nice. I'm just looking for an alternative.  

To give some context, kindly consider the following example. To
  expand $(x+a)^n$ we think of it as a combinatorics problem :
  $$(x+a)(x+a)\cdots (x+a)$$ To get $x$ term we need to
  choose $x$ from any one product and $a$ from the rest. Thus the $x$
  term would be $\binom{n}{1}xa^{n-1}$
  To get the $x^2$ term
  we need to choose $x$ from any two products and $a$ from the rest:
  $\binom{n}{2}x^2a^{n-2}$

I'm wondering if the product rule can be seen using combinatorics $$\begin{align}
(fg)^{'} &=f'g+fg'\\
(fg)^{''}&=(f'g+fg')^{'} = f''g+2f'g'+fg''
\end{align}$$
This looks almost same as the earlier problem of expanding $(x+a)^n$. I'm pretty sure these two problems are identical, but I'm not able to make the connection. Any help ?

REPLY [3 votes]: Define three operators: $D_\times$ represents differentiation of a product; $D_1$ represents differentiation of the first term of the product; and $D_2$ represents differentiation of the second term. Then the simple product rule $(fg)' = f'g + fg'$ can be written $D_\times = D_1 + D_2$. Observe that $D_1$ and $D_2$ are commutative, so you can apply the binomial theorem to $(D_1 + D_2)^n$.