Why General Leibniz rule and Newton's Binomial are so similar? The binomial expansion:
$$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$$
The General Leibniz rule (used as a generalization of the product rule for derivatives):
$$(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}$$
Both formulas can be obtained simply by induction; Newton's binomial also has a combinatorial proof (here's the relevant wikipedia page).
It's striking how these formulas are similar; is there a possible connection between them?
I was thinking that maybe the General Leibniz rule could be obtained using a combinatorial argument as well (hence the binomial coefficients)... 

Taylor series expansion of $f(x+h)$
$$
f(x+h)=f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....
$$
Taylor series expansion of $g(x+h)$
$$
g(x+h)=g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+....
$$
Taylor series expansion of $f(x+h)g(x+h)$
$$
f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+....
$$

$$
f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+....=(f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....)(g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+....)
$$
If you order $h^n$, it will give you binom expansion
$$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$ 
Because it has exactly the same coefficient of
$$
=(1+hx +\frac{h^2 x^2 }{2!}+\frac{h^3x^3 }{3!}+....)(1+hy +\frac{h^2 y^2 }{2!}+\frac{h^3y^3 }{3!}+....)=e^{hx}e^{hy}=e^{h(x+y)}
$$
$$
e^{h(x+y)}=1+h(x+y) +\frac{h^2 (x+y)^2 }{2!}+\frac{h^3(x+y)^3 }{3!}+....
$$
If you order $h^n$ of $e^{hx}e^{hy}$  and equal to $h^n$ of $e^{h(x+y)}$ , it will give you binom expansion proof.
$$\frac{(x+y)^n }{n!}=\sum_{i=0}^n  \frac{x^i y^{n-i}}{i! (n-i)!} $$ 
$$(x+y)^n=\sum_{i=0}^n  \frac{n! x^i y^{n-i}}{i! (n-i)!} $$ 
$$(x+y)^n=\sum_{i=0}^n  {n \choose i} x^i y^{n-i} $$ 
Thus it is also true
$$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$ 
