Terence Tao, Analysis I, 3e:
Lemma 5.6.6. Let $x,y \ge 0$ be non-negative reals, and let $n,m \ge 1$ be positive integers.
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(d) We have $x > y$ if and only if $x^{1/n} > y^{1/n}$.
(e) If $x > 1$, then $x^{1/k}$ is a decreasing function of $k$. (...)
Exercise 5.6.1 requires the reader to prove lemma 5.6.6.e (amongst others). The following is my attempt to do so.
By definition, $x^{1/k} := \sup \{ y \in \mathbb{R}: y \ge 0 \; \wedge \; y^k \le x \}.$
If $x > 1$, then $x^{1/k} > 1$ (proof skipped) and
$$ \tag{1}\label{eqn.lubLargerOne} \sup \{ y \in \mathbb{R}: y \ge 0 \; \wedge \; y^{k + 1} \le x \} > 1. $$
Since the least upper bound is greater than $1$, there is an element greater than $1$ in the set. Otherwise, if all elements were less than $1$ or equal to $1$, there would always be an upper bound less than the least upper bound. A contradiction.
Hence, for all $y > 1$, and $x' := xy^{-1}$,
$$ \tag{2}\label{eqn.normalization} y^{k+1} = y^k y \le x \Rightarrow y^k \le x' < x $$
such that $x > 1$ implies
$$ \tag{3}\label{eqn.identity} \sup \{ y \in \mathbb{R}: y \ge 0 \; \wedge \; y^{k + 1} \le x \} = \sup \{ y \in \mathbb{R}: y \ge 0 \; \wedge \; y^k \le x' \}. $$
From 5.6.6.d:
$$ \tag{4}\label{eqn.conclusion} x > x' \Rightarrow x^{1/k} > (x')^{1/k} = x^{1/(k+1)}, $$
which requires that $x^{1/k}$ is a decreasing function of $k$.
Correct?