Nth Derivative of Reciprocal of Mth Power

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

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

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

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \frac {d^n}{dx^n} \frac 1 {x^m} = \frac {\left({-1}\right)^n m^{\overline n}} {z^{m + n}}$

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

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

Induction Step
This is our induction step:

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