Nth Derivative of Reciprocal of Mth Power

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

The $n$th derivative of $$\frac 1 {x^m}$$ w.r.t. $x$ is:
 * $$\frac {d^n}{dx^n} \frac 1 {x^m} = \frac {\left({-1}\right)^n m^{\overline n}} {z^{m + n}}$$

where $$m^{\overline n}$$ denotes the rising factorial.

Corollary

 * $$\frac {d^n}{dx^n} \frac 1 x = \frac {\left({-1}\right)^n n!} {z^{n + 1}}$$

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:
 * $$\frac {d^n}{dx^n} \frac 1 {x^m} = \frac {\left({-1}\right)^n m^{\overline n}} {z^{m + n}}$$

Basis for the Induction
$$P(1)$$ is true, as this is the case:

$$ $$ $$

which matches the proposition as $$m^{\overline 1} = m$$ from the definition of rising factorial.

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:
 * $$\frac {d^k}{dx^k} \frac 1 {x^m} = \frac {\left({-1}\right)^k m^{\overline k}} {z^{m + k}}$$

Then we need to show:
 * $$\frac {d^{k+1}}{dx^{k+1}} \frac 1 {x^m} = \frac {\left({-1}\right)^{k+1} m^{\overline {k+1}}} {z^{m + k + 1}}$$

Induction Step
This is our induction step:

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

$$ $$ $$ $$ $$

Proof of Corollary
Follows directly by putting $$m = 1$$.