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: $$ 1 = 2L $$ $$ \frac{1}{2} = L $$