Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ I'm supposed to calculate:
$$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$
By using WolframAlpha, I might guess that the limit is $\frac{1}{2}$, which is a pretty interesting and nice result. I wonder in which ways we may approach it.
The probabilistic way: 


This is $P[N_n\leqslant n]$ where $N_n$ is a random variable with Poisson distribution of parameter $n$. Hence each $N_n$ is distributed like $X_1+\cdots+X_n$ where the random variables $(X_k)$ are independent and identically distributed with Poisson distribution of parameter $1$. 

By the central limit theorem, $Y_n=\frac1{\sqrt{n}}(X_1+\cdots+X_n-n)$ converges in distribution to a standard normal random variable $Z$, in particular, $P[Y_n\leqslant 0]\to P[Z\leqslant0]$. 

Finally, $P[Z\leqslant0]=\frac12$ and $[N_n\leqslant n]=[Y_n\leqslant 0]$ hence $P[N_n\leqslant n]\to\frac12$, QED.



The analytical way, completing your try:


  Hence, I know that what I need to do is to find $\lim\limits_{n\to\infty}I_n$, where
  $$
I_n=\frac{e^{-n}}{n!}\int_{0}^n (n-t)^ne^tdt.$$


To begin with, let $u(t)=(1-t)e^t$, then $I_n=\dfrac{e^{-n}n^n}{n!}nJ_n$ with
$$
J_n=\int_{0}^1 u(t)^n\mathrm dt.
$$
Now, $u(t)\leqslant\mathrm e^{-t^2/2}$ hence
$$
J_n\leqslant\int_0^1\mathrm e^{-nt^2/2}\mathrm dt\leqslant\int_0^\infty\mathrm e^{-nt^2/2}\mathrm dt=\sqrt{\frac{\pi}{2n}}.
$$
Likewise, the function $t\mapsto u(t)\mathrm e^{t^2/2}$ is decreasing on $t\geqslant0$ hence $u(t)\geqslant c_n\mathrm e^{-t^2/2}$ on $t\leqslant1/n^{1/4}$, with $c_n=u(1/n^{1/4})\mathrm e^{-1/(2\sqrt{n})}$, hence
$$
J_n\geqslant c_n\int_0^{1/n^{1/4}}\mathrm e^{-nt^2/2}\mathrm dt=\frac{c_n}{\sqrt{n}}\int_0^{n^{1/4}}\mathrm e^{-t^2/2}\mathrm dt=\frac{c_n}{\sqrt{n}}\sqrt{\frac{\pi}{2}}(1+o(1)).
$$
Since $c_n\to1$, all this proves that $\sqrt{n}J_n\to\sqrt{\frac\pi2}$. Stirling formula shows that the prefactor $\frac{e^{-n}n^n}{n!}$ is equivalent to $\frac1{\sqrt{2\pi n}}$. Regrouping everything, one sees that $I_n\sim\frac1{\sqrt{2\pi n}}n\sqrt{\frac\pi{2n}}=\frac12$.


  Moral:
  The probabilistic way is shorter, easier, more illuminating, and more fun.
  
  Caveat:
  My advice in these matters is, clearly, horribly biased.
On this page there is a nice collection of evidence.

I add another proof which also uses the Stirling formula.   

$\displaystyle e^{-n}\sum\limits_{k=0}^n\frac{n^k}{k!} = e^{-n}\sum\limits_{k=0}^n\frac{k^k (n-k)^{n-k}}{k!(n-k)!} \hspace{4cm}$ e.g. here

$\displaystyle \lim\limits_{n\to\infty} e^{-n}\sum\limits_{k=1}^{n-1}\frac{e^k e^{n-k}}{\sqrt{2\pi k (1+\mathcal{O}(1/k))}\sqrt{2\pi (n-k)(1+\mathcal{O}(1/(n-k)))}} $

$\displaystyle = \lim\limits_{n\to\infty} \frac{1}{2\pi}\frac{1}{n}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{\frac{k}{n}\left(1-\frac{k}{n}\right)}} =\frac{1}{2\pi} \int\limits_0^1\frac{dx}{\sqrt{x(1-x)}}=\frac{\Gamma(\frac{1}{2})^2}{2\pi~\Gamma(1)} = \frac{1}{2}$