Upper Bound on Probability of Maximum Statistic from Exponential Distribution Let $X_1,\cdots,X_n$ be I.I.D. Exponential$(\lambda)$. Let $X_{(n)}$ denote the maximum ordered statistic.
Prove that $$P\left(X_{(n)}\geq \frac{2\log(n)}{\lambda}\right)\leq\frac{1}{n}.$$
Work so far:
$$
\begin{align} 
P\left(X_{(n)}\geq\frac{2\log(n)}{\lambda}\right)&=1-F_{X_{(n)}}\left(\frac{2\log(n)}{\lambda}\right)\\&=1-\left[F_X\left(\frac{2\log(n)}{\lambda}\right)\right]^n\\&=1-\left[1-\text{exp}\{-2\log(n)\}\right]^n\\&=1-\left[1-\frac{1}{n^2}\right]^n
\end{align} 
$$
From here, I can't think of a clever way to change the RHS to be less than or equal to $\frac{1}{n}$. Any hints or tips would be much appreciated!

Hint: Bernoulli’s inequality:

$$(1+x)^n\geq 1+nx\quad x\geq -1, n\in\mathbb N.$$

