Nth Derivative of Natural Logarithm

Theorem
The $n$th derivative of $\map \ln x$ for $n \ge 1$ is:
 * $\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$

Proof
Proof by induction:

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

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

This follows by Derivative of Natural Logarithm Function.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

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

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

Induction Step
This is our induction step:

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

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