real analysis - Limit of Series with Variable Lower Bound

I'm trying to compute the following limit of a series: $$\lim_{n\to\infty} \sum_{k = n+1}^{\infty}\frac{e^{-n}n^{k}}{k!}$$

Factoring $e^{-n}$ out of the sum, applying the definition of $e^{x}$ as a power series and then simplifying gives me that this limit is equivalent to: $$1 - \lim_{n\to\infty}(e^{-n}\sum_{k = 0}^{n}\frac{n^{k}}{k!})$$

This limit certainly seems easier to work with as the lower index is now fixed but the coupling of $n$ in both the upper index and summand is still troubling me. Looking at the partial sums it's clear that this evaluates to $\frac{1}{2}$ but showing it is a whole different story. I've been trying to find a chain of inequalities that will give me this result via the Squeeze Theorem. I've observed:

$$\sum_{k = 0}^{n}\frac{n^{k}}{n!} \leq \sum_{k = 0}^{n}\frac{n^{k}}{k!} \leq \sum_{k = 0}^{\infty}\frac{n^{k}}{k!} = e^n \space \space \forall n \in \mathbb{N}$$

$$\implies e^{-n}\sum_{k = 0}^{n}\frac{n^{k}}{n!} \leq e^{-n}\sum_{k = 0}^{n}\frac{n^{k}}{k!} \leq 1 \space \space \forall n \in \mathbb{N}$$

But this does me no good right now. Looking on Wolfram I see that $e^{-n}\sum_{k = 0}^{n}\frac{n^{k}}{k!} = \frac{\Gamma{(n+1, n)}}{n\Gamma(n)}$ where $\Gamma{(-,-)}$ is the incomplete Gamma function but using this seems like the wrong way to go about it as I was told an elegant solution exists. Any hints or suggestions are greatly appreciated. Thanks!