Nth Derivative of Reciprocal of Mth Power

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

The $n$th derivative of $\dfrac 1 {x^m}$ $x$ is:
 * $\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$

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

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * $\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$

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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\dfrac {\d^k} {\d x^k} \dfrac 1 {x^m} = \dfrac {\paren {-1}^k m^{\overline k} } {x^{m + k} }$

Then we need to show:
 * $\dfrac {\d^{k + 1} } {\d x^{k + 1} } \dfrac 1 {x^m} = \dfrac {\paren {-1}^{k + 1} m^{\overline {k + 1} } } {x^{m + k + 1} }$

Induction Step
This is our induction step:

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