Relationship between factorial and derivatives I was wondering if there is any relationship between factorials and derivatives because I notice that if we had $x^n$ and we take the $n$-th derivative of this function it will be equal to the factorial of $n$: $$\frac{d^n}{dx^n}(x^n)=n!$$
 A: Let $U_{n,x}=\{(x_1,x_2,\dots,x_n)\mid 0<x_1<x_2<\cdots <x_n\leq x\}$. It is not hard to show, by a symmetry argument, that the hypervolume of this region is $\frac{x^n}{n!}$, because there are $n!$ ways to permute the $x_i$ to get a different order, and this covers "almost all" of the $n$-dimension hypercube with side $x$.
An interesting feature can be seen that the value:
$$|U_{x+y,n}| = \sum_{k=0}^{n} |U_{x,k}|\cdot|U_{y,n-k}|$$
This is the binomial theorem written geometrically. It's saying that we can break up $U_{x+y,n}$ into components $A_k$ where the first $k$ elements of $(x_1,\dots,x_n)$ are less than $x$ and the next $n-k$ elements are greater than $x$. (Where $|U_{z,0}|$ is considered to be $1$.)
This lets us calculate the derivative, because $y=|U_{y,1}|$ and you get:
$$\frac{U_{x+y,n}-U_{x,n}}{y} = |U_{x,n-1}| + \sum_{k=0}^{n-2} |U_{x,k}|\cdot \frac{|U_{y,n-k}|}{y}$$ and we get that $\frac{|U_{y,j}|}{y}\to 0$ as $y\to 0$ when $j>1$.
It feels like there is something more that can be done with this idea, perhaps using something like:
$$f^{(n)}(x)=\lim_{h_1\to 0} \lim_{h_2\to 0} \cdots \lim_{h_n\to 0}\text{ some horrible expression }$$
A: Yes, and that's precisely why $n!$ appears in the denominator of the term of a Taylor series containing $x^n$ (for simplicity, I'll assume the series is centered at $x=0$).
That term is $\frac{f^{(n)}(0)}{n!}x^n$. When you take $n$ derivatives and plug in $x=0$, you get just $f^{(n)}(0)$ as desired. That's why the Taylor series has the correct derivatives of all orders at $x=0$; all other terms vanish (they have either been differentiated away or they still contain a real factor of $x$), leaving just exactly what you want.
A: Look at the general expression $$y_i = \frac{{\rm d}^i x^n}{{\rm d}x^i}$$
This is the i-th derivative of $x^n$ for $i=1\ldots n$. This can be directly evaluated as
$$ \boxed{ \frac{{\rm d}^i x^n}{{\rm d}x^i} = \frac{n!}{(n-i)!} x^{n-i} } $$
So the i-th derivative of a n-th order polynomial contains terms of $n!$, $(n-1)!$, $(n-2)!$ etc
