Well, we can just get rid of $e^{-n}$ rather easily, but that's not what we should do.
$ \lim_{n\rightarrow\infty} e^{-n} \sum_{i=0}^n \frac{n^i}{i!} $
There's something called the Incomplete Gamma Function. It satisfies:
$ \frac{\Gamma(n+1, n)}{n! e^{-n}} = \sum_{k=0}^n \frac{n^k}{k!}$
Substitute:
$ \lim_{n\rightarrow\infty} e^{-n} \frac{\Gamma(n+1, n)}{n! e^{-n}} $
Get rid of $e^{-n}$: $ \lim_{n\rightarrow\infty} \frac{\Gamma(n+1, n)}{n!} $
Now what? Well make a substitution: $ \lim_{n\rightarrow\infty} \frac{\Gamma(n+1, n)}{\Gamma(n+1)} = \frac{1}{2}$
(Note that the following proof might be incorrect, although my CAS agrees with the result and I think it is.)
In order to show this, there is an identity that $\Gamma(a, x) + \gamma(a, x) = \Gamma(a) $, so $\Gamma(a, x) = \Gamma(a) - \gamma(a, x)$. Now find: $ \lim_{n\rightarrow\infty} \frac{\Gamma(n+1) - \Gamma(n+1,x)}{\Gamma(n+1)} $
$ 1 - \lim_{n\rightarrow\infty} \frac{\Gamma(n+1,x)}{\Gamma(n+1)} $
But this is the same as our other limit. If we have: $ \lim_{n\rightarrow\infty} \frac{\Gamma(n+1,x)}{\Gamma(n+1)} = L $
Then: $ 1 - L = L $
So: $ 1 = 2L $ $ \frac{1}{2} = L $