Why is $ \lim_{n \to \infty} \left(\frac{n-1}{n+1} \right)^{2n+4} = \frac{1}{e^4}$ According to WolframAlpha, the limit of
$$ \lim_{n \to \infty} \left(\frac{n-1}{n+1} \right)^{2n+4} = \frac{1}{e^4}$$
and I wonder how this result is obtained.
My approach would be to divide both nominator and denominator by $n$, yielding
$$ \lim_{n \to \infty} \left(\frac{1-\frac{1}{n}}{1 + \frac{1}{n}} \right)^{2n+4} $$
As $ \frac{1}{n} \to 0 $ as $ n \to \infty$, what remains is 
$$ \lim_{n \to \infty} \left(\frac{1-0}{1 + 0} \right)^{2n+4} = 1 $$
What's wrong with my approach?

$$ \lim_{n \to \infty} \left(\frac{n-1}{n+1} \right)^{2n+4} $$
$$ =\lim_{n \to \infty} \left(\left(1+\frac{(-2)}{n+1} \right)^{\frac{n+1}{-2}}\right)^{\frac{-2(2n+1)}{n+1}}$$
$$ = \left(\lim_{n \to \infty}\left(1+\frac{(-2)}{n+1} \right)^{\frac{n+1}{-2}}\right)^{\lim_{n \to \infty}\left(\frac{-4-\frac2n}{1+\frac1n}\right)}$$
$$=(e)^{-4}\text{ as } n \to \infty, \frac{n+1}{-2}\to-\infty\text{ and } \lim_{m\to\infty}\left(1+\frac1m\right)^m=e$$
