Differentiation of the term x^n n times. Kindly verify the proof i couldn't find this anywhere. I am fairly new to differentiation so i apologize for mistakes if any... 
$$\frac{d}{dx}\left(x^n\right)=nx^{n-1}$$
$$\frac{d}{dx}\left(nx^{n-1}\right)=n\left(n-1\right)x^{n-1-1}=n\left(n-1\right)x^{n-2}$$
Hence if differentiated $n$ times:
$$\frac{d^n}{dx^n}\left(x^{n}\right)=n!x^{n-1-1-1-...-1}=n!x^{n-n}$$
$$\frac{d^n}{dx^n}\left(x^{n}\right)=n!$$
Thank you
 A: welcome on SE, you should use TeX to write formulae.
Anyway, the identity $(\frac{d}{dx})^nx^n=n!$ is correct, as well as the way you derived it, but to have a formal proof you probably have to use induction.
Let's suppose this identity holds for $n$ (and you can immediately verify that it holds for $n=1$), then, for $n+1$ we have:
$(\frac{d}{dx})^{n+1}x^{n+1}=(\frac{d}{dx})^{n}(n+1)x^n$
but we already know the n-th derivative of $x^n$ (the $n+1$ in front of it is just a constant) so we can write
$(\frac{d}{dx})^{n+1}x^{n+1}=(n+1)n!=(n+1)!$
A: We can do a proof by induction so we start by proving true for n = 1
so we consider x
$\partial_xx = 1 = 1!$ where $\partial_x^n$ is the nth partial derivative with respect to x i.e $ \frac {d^n}{dx^n}$
now we assume true for n
$\partial_x^nx^n=n!$
now we prove true for n+1
$\partial _x ^ {n+1} x ^ {n+1} = \partial_x^n(n+1)\cdot x^n = (n+1)\cdot \partial_x^nx^n$
but we know from our assumption that:
$\partial_x^nx^n=n!$
so if we plug that in we get
$\partial _x ^ {n+1} x ^ {n+1} = n!\cdot(n+1)= (n+1)!$
which is what we wanted
A: The proof should be by induction; the base case, $n=1$, is obvious: $Dx=1=1!$.
Now suppose you know that $D^{n}x^{n}=n!$ and try computing
$$
D^{n+1}x^{n+1}=D^n(Dx^{n+1})
$$
(using linearity of derivation).
