Mathematical Prove (1)

Prove $(1-x)^{x^{-1}N} < e^{-N}$ for $x \in (0,1)$

Assume $(1-x)^{x^{-1}N} < e^{-N}$, then
$ln((1-x)^{x^{-1}N}) < -N$
$x^{-1}Nln(1-x) < -N$
$x^{-1} ln(1-x) < -1$
$ln(1-x) < -x$
$ln(1-x) + x < 0$

So we can prove $ln(1-x) + x < 0$ instead
Let $f(x) = ln(1-x) + x$,
$f'(x) = \frac{d}{d(1-x)}ln(1-x)\cdot\frac{d}{dx}(1-x)+\frac{dx}{dx} = -\frac{1}{1-x} + 1$

Since $f'(x) < 0$ for $x \in (0,1)$ and $f'(x) = 0$ when $x = 0$, $f(x)$ is strictly decreasing for $x \in (0,1)$ with the greatest value $f(0) = ln(1-0) + 0 = 0$. Therefore, $f(x) < 0$ for $x \in (0,1)$ and so $(1-x)^{x^{-1}N} < e^{-N}$ for $x \in (0,1)$