algebra precalculus - Proving an inequality $log$ and $e$

This inequality should be fairly easy to show. I think I'm just having trouble looking at it the right way (It's used in a proof without explanation).

$$(1-\frac{1}{\log ^{2} n})^{(2 \log n) -1}\geq e^{-2/\log n}$$

Any help is much appreciated. Thanks
Edit: Log is base 2

Answer

Assuming $n>2$ are natural numbers and $\log = \log_2$:

$(1 + \frac 1n)^n$ is increasing, $\lim_{n\to \infty}(1+ \frac 1n)^n =e $ and $(1+\frac 1x)^x < e$ for $x \ge 1$.

And $(1-\frac 1x)^x > \frac 1e$ for $x \ge 1$.

So $ (1 - \frac 1{\log^2 n})^{\log^2 n} > e^{-1}$

$( 1 - \frac 1{\log^2 n})^{2\log n} > e^{\frac{-2}{\log n}}$

$( 1 - \frac 1{\log^2 n})^{2\log n - 1} > \frac {e^{\frac{-2}{\log n}}}{1 - \frac 1{\log^2 n}}$

If $n > 2$ and $\log n = \log_2 n > 1$ then $0< {1 - \frac 1{\log^2 n}} < 1$ and

$( 1 - \frac 1{\log^2 n})^{2\log n - 1} > \frac {e^{\frac{-2}{\log n}}}{1 - \frac 1{\log^2 n}}>e^{\frac{-2}{\log n}} $

If $n = 2$ then

$( 1 - \frac 1{\log^2 n})^{2\log n - 1} =$

$( 1 - \frac 1{\log^2 2})^{2\log 2 - 1} = 0^0$ is undefined.

Likewise if $n=1$ we have division by $0$.

Perhaps $\log = \ln =\log_e$?