Evaluate $ \lim n \left[ 1-\frac{(n+1)^n}{en^n}\right] $ Evaluate the following limit of sequence

$$\lim_{n\to +\infty} n \left[ 1-\frac{(n+1)^n}{en^n}\right] $$

I've transformed it in a 0/0 inequality and tried to apply L'Hospital one time, but the function seems even more complicated.

Thanks in advance!
Since (as long as $n\to +\infty$):
$$\left(1+\frac{1}{n}\right)^n = \exp\left(n\log\left(1+\frac{1}{n}\right)\right)=\exp\left(1-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right) $$
we have:
$$1-\frac{1}{e}\left(1+\frac{1}{n}\right)^n=1-\exp\left(-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right)=\frac{1}{2n}+O\left(\frac{1}{n^2}\right)$$
hence:
$$\lim_{n\to +\infty} n\cdot\left[1-\frac{1}{e}\left(1+\frac{1}{n}\right)^n\right]=\frac{1}{2}.$$
We just used the Taylor series of $\log(1+z)$ and $e^z$ around $z=0$.
Use the fact that $lim_{n \rightarrow  \infty} (1+1/n)^n = e$ to get limit of your sequence equal to $1-1/e^2$