I'm reading Introduction to Random Graphs by Frieze and Karonski. In chapter 2, at the proof of Lemma 2.12, a random variable $X_k$ is introduced, which counts isolated trees of order $k$ in the random graph with $p=\frac{c}{n}$, $c<1$. Its expectation equals $$\mathbb{E}(X_k)=\binom{n}{k}k^{k-2}p^{k-1}(1-p)^{k(n-k)+\binom{k}{2}-k+1}$$ (equation (2.8) in the book). Later on (page 32) it is bounded as follows:
$$\mathbb{E}(X_k)\leq \frac{A}{\sqrt{k}}\left(\frac{ne}{k}\right)^{k}k^{k-2}\left(\frac{c}{n}\right)^{k-1}\left(1-\frac{k}{2n}\right)^{k-1}e^{-ck+\frac{ck^{2}}{2n}}$$ where $A$ is some positive constant. Note that it is only known that $k\leq n$.
I do not entirely understand that bound; specifically where does the $\left(1-\frac{k}{2n}\right)^{k-1}$ term come from.
I can see that the bound $\binom{n}{k}\leq \frac{1}{2\sqrt{k}}\left(\frac{ne}{k}\right)^{k}$ was used on the binomial coefficient. I believe that $e^{-ck+\frac{ck^{2}}{2n}}$ comes from bounding $\left( 1-\frac{c}{n} \right)^{kn-\frac{k^2}{2}}$. However the remaining $\left( 1-\frac{c}{n} \right)^{-\frac{3k}{2}+1}$ definitely cannot be bounded by $\left(1-\frac{k}{2n}\right)^{k-1}$, as it is greater than $1$. So how did that term get there?
Unfortunately, this is not explained at all in the text. Here is where that term comes from.
To get from $\binom nk$ to $\frac{A}{\sqrt k} (\frac{ne}{k})^k$, we first use $\binom nk \le \frac{n^k}{k!}$, then $k! \ge \frac{A}{\sqrt k} (\frac ke)^k$. However, when $k$ is large, we give a lot away in the first bound $\binom nk \le \frac{n^k}{k!}$. More precisely, we have $$ \binom nk = \frac{n^k}{k!} \cdot \frac{(n-1)(n-2)\dotsm (n-k+1)}{n^{k-1}}. $$ Now let's look at this factor some more. By AM-GM, the product $(n-1)(n-2)\dotsm (n-k+1)$ is at most $\left(\frac{(n-1) + (n-2) + \dots + (n-k+1)}{k-1}\right)^{k-1}$, or $(n - \frac k2)^{k-1}$, giving us an upper bound of exactly $$ \frac{(n - \frac k2)^{k-1}}{n^{k-1}} = \left(1 - \frac{k}{2n}\right)^{k-1}. $$ In case you are also worried about the remaining factor of $(1 - \frac cn)^{-3k/2 + 1}$, it is at most $\exp(\frac{3ck}{2n} - \frac cn) \le e^{3c/2}$, which is absorbed in the constant $A$.