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:
 * $$\frac {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

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

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:
 * $$\frac {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:
 * $$\frac d {dx} x^m = m x^{m-1}$$

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

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:
 * $$\frac {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:
 * $$\frac {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:
 * $$\frac {d^k}{dx^k} x^m = m^{\underline k} x^0 = m!$$

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

Then $$\frac {d^{k+1}}{dx^{k+1}} = 0$$ 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:
 * $$\frac {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.