Bound for $(1-\frac{1}{n})^t$

Question

I'm having trouble proving that:

For any constant $\epsilon > 0$ and $n > 1$:

$$ \left(1-\frac{1}{n}\right)^{n lg\left(n^{\epsilon}\right)} \leq \frac{1}{n^{\epsilon}}$$

I'm using $lg(n)$ as $log_2(n)$. Any help is appreciated, hints for critical points of the proof are welcome as well.

Answer 1

Hint :

The $\ln$ function is strictly increasing and defined for numbers $>0$. Since these hold in your expression, it is :

$$\bigg(1-\frac{1}{n}\bigg)^{n\ln{n^\epsilon}} \leq \frac{1}{n^\epsilon} \Rightarrow \ln\bigg(1-\frac{1}{n}\bigg)^{n\ln{n^\epsilon}} \leq \ln\bigg(\frac{1}{n^\epsilon}\bigg)$$

$$\Leftrightarrow$$

$$n\cdot \epsilon \ln n \cdot \ln\bigg(1-\frac{1}{n}\bigg)\leq -\epsilon\ln n$$

$$\Leftrightarrow$$

$$\bigg[n\ln\bigg(1-\frac{1}{n}\bigg)-1\bigg]\epsilon\ln n \leq 0$$