Nth Derivative of Nth Power

Corollary to Nth Derivative of Mth Power
Let $n \in \Z$ be an integer such that $n \ge 0$.

The $n$th derivative of $x^n$ w.r.t. $x$ is:
 * $\dfrac {\mathrm d^n}{\mathrm d x^n} x^n = n!$

where $n!$ denotes $n$ factorial.

Proof
From Nth Derivative of Mth Power, we have:


 * $\dfrac {\mathrm d^n}{\mathrm d x^n} x^m = \begin{cases}

m^{\underline n} x^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$ where $m^{\underline n}$ denotes the falling factorial.

Putting $m = n$:
 * $\dfrac {\mathrm d^n}{\mathrm d x^n} x^n = n^{\underline n}$

where from the definition of the falling factorial:
 * $n^{\underline n} = n!$

Hence the result.