Prove that $ \left(1-\frac{1}{n}\right)^n > \frac{1}{6} $ for $n \in \mathbb{N}$, $ n\ge 2$

Indeed, the affirmation is true even if $n$ is not a natural ($ n\geq 2 $ ) and we can prove it using calculus. But, this is part of a question about limit of sequences (that is why I fixed $n$ a natural number), so I can't officially use derivative yet. Then, I was thinking to use induction.

For $n = 2$ it is clearly true. If we have $ (1 - \frac{1}{n})^n > \frac{1}{6}$ , then for $n>1$ 
$$ \left(1-\frac{1}{n+1}\right)^{n+1} > \left(1-\frac{1}{n}\right)^{n+1} > \frac{1}{6} \left(1-\frac{1}{n} \right) .$$ But it is not helping.

I would appreciate some help.

Thanks!