Logarithm of Power/Natural Logarithm/Natural Power

Theorem
Let $x \in \R$ be a strictly positive real number.

Let $n \in \R$ be any natural number.

Let $\ln x$ be the natural logarithm of $x$.

Then:
 * $\ln \left({x^n}\right) = n \ln x$

Proof
Proof by Mathematical Induction:

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\forall x \in \R_{>0}: \ln \left({ x^n }\right) = n\ln \left({ x }\right)$

Basis for the Induction
$P \left({0}\right)$ is the case:
 * $\forall x \in \R_{>0}: \ln \left({ 1 }\right) = 0 $

from Logarithm of 1 is 0

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 0 $, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:
 * $\forall x \in \R_{>0}: \ln \left({ x^k }\right) = k\ln \left({ x }\right)$

Then we need to show:
 * $\forall x \in \R_{>0}: \ln \left({ x^{k+1} }\right) = \left({ k + 1 }\right) \ln \left({ x }\right)$

Inductive Step
This is our induction step:

Fix $x \in \R_{>0}$.

Then:

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

Therefore:
 * $\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: \ln \left({ x^n }\right) = n\ln \left({ x }\right)$