Nth Derivative of Natural Logarithm

Theorem
The $n$th derivative of $\ln(x)$ for $n≥1$ is:
 * $\dfrac {\mathrm d^n}{\mathrm d x^n} \ln(x)= \dfrac{(n-1)!(-1)^{n-1}}{x^n} $

Proof
Proof by induction:

For all $n \in \N_{>1}$, let $P \left({n}\right)$ be the proposition:
 * $\dfrac {\mathrm d^n}{\mathrm d x^n} \ln(x)= \dfrac{(n-1)!(-1)^{n-1}}{x^n} $

Basis for the Induction
$P(1)$ is true, as this just says:
 * $\dfrac {\mathrm d} {\mathrm dx} \ln(x) = \dfrac {1}{x}$

This follows by Derivative of Natural Logarithm Function

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:
 * $\dfrac {\mathrm d^k}{\mathrm d x^k} \ln(x)= \dfrac{(k-1)!(-1)^{k-1}}{x^k} $

Then we need to show:
 * $\dfrac {\mathrm d^{k+1}}{\mathrm d x^{k+1}} \ln(x)= \dfrac{k!(-1)^{k}}{x^{k+1}} $

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\dfrac {\mathrm d^n}{\mathrm d x^n} \ln(x)= \dfrac{(n-1)!(-1)^{n-1}}{x^n} $