Here is how $(1)$ is obtained in the regime where $k$ is small compared to $n$. First, take logs, and use $\log(1-x)=-x+O(x^2)$.
\begin{align}
\log\frac{n!}{(n-k)!n^k}
&= \sum_{i=0}^{k-1}\log\left(1-\frac in\right)
\\&= \sum_{i=0}^{k-1}\left(-\frac{i}{n}+O\Big(\frac{i^2}{n^2}\Big)\right)
\\&= -\frac{k(k-1)}{2n}+O\Big(\frac {k^3}{n^2}\Big)
\end{align}
Exponentiating, we get
$$
\frac{n!}{(n-k)!n^k}\approx \exp\left(\frac{-k^2}{2n}+O\Big(\frac {k^3}{n^2}\Big)\right)
$$
In the regime where $k=o(n^{2/3})$, the above becomes $\exp(-k^2/2n +o(1))$, which means that the approximation of $\exp(-k^2/2n)$ is good as $n\to\infty$.
Suppose instead that $k$ is large, by which I mean $k\ge Cn^{2/3}$. Here, we need to be a little more careful. We use the inequality $\log(1-x)\le -x$ to get an inequality for $n!/(n-k)!n^k$, instead of a big-Oh approximation. The result is
$$
\frac{n!}{(n-k)!n^k}\le \exp\left(-\frac{k(k-1)}{2n}\right) \approx \exp\left(-\frac{k^2}{2n}\right)\le \exp\left(-\frac{(Cn^{2/3})^2}{2n}\right) =\exp(-(C^2/2)n^{1/3})
$$
This shows that both the exact answer and the approximation are approximately zero, which means that the approximation is good.
I finally figured out your error. The problem is this step. You claim that
$$
\left(1-\frac{k}{n}\right)^{-(n-k+1/2)}=\left[\left(1-\frac1{n/k}\right)^{-n/k}\right]^{\frac{n-k+1/2}{n/k}} \stackrel{\color{red}{\text{error}}}{\sim} e^{\frac{n-k+1/2}{n/k}}
$$
You seem to using the following false lemma:
False Lemma: Let $a_n,b_n,c_n$ be positive sequences. Suppose that $a_n\sim b_n$ as $n\to \infty$. Then $(a_n)^{c_n} \sim (b_n)^{c_n}$ as $n\to\infty$.
You are applying this with $a_n=(1-\frac1{n/k})^{-n/k}$, $b_n=e$, and $c_n=\frac{n-k+1/2}{n/k}$, where we are thinking of $k=k(n)$ as a function of $n$.
To see why this lemma is false, take $a_n=1+1/n$, $b_n=1$, and $c_n=n$.
