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 {\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.

Proof
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\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}$

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

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

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 {\mathrm d^k}{\mathrm d x^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 {\mathrm d^{k+1}}{\mathrm d x^{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:
 * $\dfrac {\mathrm d^k}{\mathrm d x^k} x^m = m^{\underline k} x^0 = m!$

Then $\dfrac {\mathrm d^{k+1}}{\mathrm d x^{k+1}} x^m = \dfrac {\mathrm d}{\mathrm d x} m! = 0$ 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:
 * $\dfrac {\mathrm d^k}{\mathrm d x^k} x^m = 0$.

Then $\dfrac {\mathrm d^{k+1}}{\mathrm d x^{k+1}} = 0$ 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 {\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}$