If $\Pr(S_n=k)=e^{-n}\frac{n^k}{k!}$ then as $\frac{(k-n)}{\sqrt n}\to x$, show that $\sqrt{2\pi n}\Pr(S_n=k)\to\exp(-x^2/2)$
$e^{-n}\frac{n^k}{k!}=e^{-n}\frac{n^n}{n!}\times n^{k-n}\frac{n!}{k!}\tag1$
according to Stirling's Formula, the first term tends to $\frac1{\sqrt{2\pi n}}$
so the second must be equal to $\exp(-x^2/2)$
Using Stirling again;
$n^{k-n}\frac{n!}{k!}=\dfrac{n^n/e^n\sqrt{2\pi n}}{k^k/e^k\sqrt{2\pi k}}=\sqrt{n/k}\ e^{k-n}(\frac nk)^k$
$(\star)$ and $(\frac nk)^k=\frac{n}{x\sqrt n +n}^{x\sqrt n +n}=(1-\frac{x\sqrt n}{x\sqrt n +n})^{x\sqrt n +n}\to e^{-x\sqrt n}=e^{-(k-n)}$
and it gives finally $\sqrt{n/k}*e^{k-n}(\frac nk)^k=\sqrt{n/k}*e^{k-n}e^{-(k-n)}=\sqrt{\frac nk}$
so what did go wrong, maybe $(\star)$ is not correct ?
I think, I cannot just separate the product in $(1)$ and assume that the first term converges to $\frac1{\sqrt{2\pi n}}$ as n goes to $\infty$, since if $n\to\infty$ the whole formula doesn't make sense, but can you suggest another way ?
You were slightly sloppy earlier: Stirling says that $\sqrt{2 \pi n} e^n \frac{n^n}{n!}$ (with the leading factor) tends to $1$. Without that leading factor you can't really say "tends to", you should say "is asymptotic to", since the actual limit would be zero.
As for the rest of the problem, you should first notice that $k$ must be a function of $n$ for your limit to make sense. With that in mind, first notice that the case $x=0$ is basically just Stirling's approximation. (A little bit of extra work is needed since $k$ needn't be exactly $n$ in this situation, but it's not a big deal.)
So now assume $x>0$ and thus $k \geq n$. (The latter is no real requirement, since it must be true eventually if $x> 0$.) The argument is similar when $x<0$.
After you cancel out the top factorial with part of the bottom factorial, you are left with a product of $k-n$ terms divided by a product of $k-n$ terms. Specifically:
$$\frac{n^{k-n}}{\prod_{i=1}^{k-n} (n+i)}.$$
Cancelling factors of $n$ gives
$$\frac{1}{\prod_{i=1}^{k-n} (1+i/n)}.$$
Take the log of this and see what you can do. (It's a nice exercise.)