Prove that $(fg)^{(n)} = \sum_{k=0}^n \binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$ 
Assume $f$ and $g$ are differentiable at $x$. Prove that $(fg)^{(n)} = \sum_{k=0}^n \binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$

I am assuming here $fg = f(x) g(x)$. Then we can prove this via induction. If $n = 0$ we have $1 = 1$ which is true. Now assume it is true for some $m$ we need to show it is possible for $m+1$. We have $(fg)^m = \sum_{k=0}^m f^{(k)}(x)g^{m-k}(x)$. How do we show that?
 A: \begin{align*}
(fg)^{m+1}(x) & = \left(\sum_{k=0}^m \binom{m}{k} f^{(k)}(x)g^{(m-k)}(x)\right)'\\
& = \left(\sum_{k=0}^m \binom{m}{k} f^{(k+1)}(x)g^{(m-k)}(x)\right)+\left(\sum_{k=0}^m \binom{m}{k} f^{(k)}(x)g^{(m-k+1)}(x)\right)\\
& = f^{(m+1)}(x)g(x) + \left(\sum_{k=0}^{m-1} \binom{m}{k} f^{(k+1)}(x)g^{(m-k)}(x)\right)+\left(\sum_{k=1}^{m} \binom{m}{k} f^{(k)}(x)g^{(m-k+1)}(x)\right) + f(x)g^{(m+1)}(x)\\
& = f^{(m+1)}(x)g(x) + \left(\sum_{k=0}^{m-1} \binom{m}{k} f^{(k+1)}(x)g^{(m-k)}(x)\right)+\left(\sum_{k=0}^{m-1} \binom{m}{k+1} f^{(k+1)}(x)g^{(m-k)}(x)\right) + f(x)g^{(m+1)}(x).
\end{align*}
And to finish, use the property $$\binom{m}{k-1}+ \binom{m}{k} = \binom{m+1}{k}.$$
A: This is one instance where non-standard analysis helps understanding. In crude words, you work with infinitesimals, and standardize results (with the operator  $\operatorname{st}$) to "real" objects.
Indeed, this goes back to basics, as Leibniz is one of the Godfathers of infinitesimals. Basically, the formula is similar to that of the binomial theorem, and a little investment, stuff can be cast to polynomial-like calculus. Here, this goes like this (only giving a sketch) for the first level:
$$\frac{d(fg)}{dx}=\operatorname{st}\left(\frac{(f + \mathrm df)(g + \mathrm dg) - fg}{\mathrm dx}\right) = \operatorname{st}\left(\frac{fg + f \cdot \mathrm dg + g \cdot \mathrm df + \mathrm dg \cdot \mathrm df -fg}{\mathrm dx}\right) ={f}\frac{dg}{dx} + {g}\frac{df}{dx}\,.$$
The idea is that standardization simplifies quantities. From that, you can apply induction and known results on polynomials.
A: *

*Clue 1: $(fg)^{(m+1)}=((fg)^{(m)})^{(1)}$

*Clue 2: use induction hypothesis

*Clue 3: Collect terms. Use binomial coefficient properties

