Nth Derivative of Mth Power

Theorem
Let $m \in \Z$ be an integer such that $m \ge 0$.

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

Corollary

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

where $n!$ denotes $n$ factorial.

Proof of Main Result
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\dfrac {d^n}{dx^n} x^m = \begin{cases}

m^{\underline n} x^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$.

Basis for the Induction

 * $P(1)$ is true, as this just says:
 * $\dfrac d {dx} x^m = m x^{m-1}$

This follows by Power Rule for Derivatives, which also includes the case:
 * $\dfrac d {dx} x^0 = 0$

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\dfrac {d^k}{dx^k} x^m = \begin{cases}

m^{\underline k} x^{m - k} & : k \le m \\ 0 & : k > m \end{cases}$.

Then we need to show:
 * $\dfrac {d^{k+1}}{dx^{k+1}} x^m = \begin{cases}

m^{\underline {k+1}} x^{m - \left({k+1}\right)} & : {k+1} \le m \\ 0 & : {k+1} > m \end{cases}$.

Induction Step
This is our induction step:

First, let $k < m$. Then we have:

At this stage, as $k < m$, we have that $k+1 \le m$. So far so good.

Now suppose that $k = m$. Then by the induction hypothesis:
 * $\dfrac {d^k}{dx^k} x^m = m^{\underline k} x^0 = m!$

Then $\dfrac {d^{k+1}}{dx^{k+1}} x^m = \dfrac {d}{dx} m! = 0$ by Differentiation of a Constant.

At this stage, as $k = m$, we have that $k+1 > m$. Again, so far so good.

Finally, suppose that $k > m$. Then by the induction hypothesis:
 * $\dfrac {d^k}{dx^k} x^m = 0$.

Then $\dfrac {d^{k+1}}{dx^{k+1}} = 0$ by Differentiation of a Constant.

At this stage, as $k > m$, we have that $k+1 > m$. Which is all we need.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\dfrac {d^n}{dx^n} x^m = \begin{cases}

m^{\underline n} x^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$

Proof of Corollary
Follows directly by putting $m = n$.

The specific instance has been covered in the proof of the main result above.