Defining Sequence of Natural Logarithm is Strictly Decreasing

Theorem
Let $x \in \R$ be a real number such that $x > 0$.

Let $\sequence {f_n}$ be the sequence of mappings $f_n : \R_{>0} \to \R$ defined as:
 * $\map {f_n} x = n \paren {\sqrt [n] x - 1}$

Then $\forall x \in \R_{>0}: \sequence {\map {f_n} x}$ is strictly decreasing.

Proof
Fix $t \in R_{>0}$.

Then:

Thus:
 * $n \paren {t^{n + 1} - 1} > \paren {n + 1} \paren {t^n - 1}$

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

From Power of Positive Real Number is Positive: Rational Number:
 * $\forall n \in \N : x^{1 / \paren {n \paren {n + 1} } } \in \R_{>0}$

Thus:

Hence the result, by definition of strictly decreasing sequence.