How to prove that for every integer $k \geq 2$ we have $k^{1/k} \leq e^{1/e}$ without using the first derivative test? I just stumbled across this cool property, by doing some calculus I could prove it, the function $f(x) = x^{1/x}$ has a local maximum at $x = e$ and the derivative changes sign at that point, but I was wondering if there was other were to prove it without using the first derivative test ?

Here's a proof I saw a number of years ago
that $e^{1/e} > x^{1/x}$
for $x > e$.
All it uses is
$e^z > 1+z$ for $z > 0$.
If $x > e$ then
$e^{\dfrac{x-e}{e}}
\gt 1+\dfrac{x-e}{e}
= 1+\dfrac{x}{e}-1
= \dfrac{x}{e}
$
so
$e^{\dfrac{x}{e}-1}
\gt \dfrac{x}{e}
$
or
$e^{\dfrac{x}{e}}
\gt x
$
or
$e^{1/e}
\gt x^{1/x}
$.
The case $k = 2$ is
$2^{1/2} < e^{1/e}
$
or
$\ln(2) < 2/e$
and
$\ln(2) < .7$
and
$2/e > .73$.
