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 $x^n$ with respect to $x$ is:

$\dfrac {\d^n} {\d x^n} x^n = n!$

where $n!$ denotes $n$ factorial.

Proof

From Nth Derivative of Mth Power, we have:

$\dfrac {\d^n} {\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 $m = n$:

$\dfrac {\d^n} {\d x^n} x^n = n^\underline n$

where from the definition of the falling factorial:

$n^\underline n = n!$

Hence the result.

$\blacksquare$