Why is $\lim\limits_{n\to \infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ not equal to $1$? So I saw the limit $\lim\limits_{n\to \infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ here the other day:
Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
and when I saw it, I right away thought the answer is $1$ because I thought $\lim\limits_{n\to \infty} \sum_{k=0}^n \frac{n^k}{k!} = \lim\limits_{n\to \infty} e^n$ given that $e^x = \lim\limits_{n\to \infty} \sum_{k=0}^n \frac{x^k}{k!}$ and so the result would be $\lim\limits_{n\to \infty} e^{-n}e^n = 1$ but the result is $\frac{1}{2}$, found using methods that I'm not familiar with.
Could someone please explain why my method is wrong?
Thank you so much in advance!

If the inner sum were added for k = 0 to infinity, you would be right. But it is only added for k = 0 to n. It seems that if you add the first (n+1) terms of the sum up to n, you get about half of the complete infinite sum (which you would need to prove), that is about $e^n / 2$. 
