$\lim_{n\to \infty}\left(1 - \frac {1}{n^2}\right)^n =?$ Can you give any idea regarding the evaluation of the following limit? 
$\lim_{n\to \infty}\left(1 - \frac {1}{n^2}\right)^n$
We know that $\lim_{n\to \infty}\left(1 - \frac {1}{n}\right)^n = e^{-1}$, but how do I use that here?

General approach for ${(1^{\pm})}^{\infty}$ : 
$$f^g=e^{g\ln f}=e^{g \ln(1+f-1)}=e^{g(f-1)}$$
where I have used $\ln(1+x)=x-x^2/2...$ I have ignored other small terms.
Here, $f=1-\frac 1 {n^2}$ and $g=n$
