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.

Corollary

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

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

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

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

Induction Step
This is our induction step:

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

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