In this proof from "The Probabilistic Method" by Alon and Spencer, p.206, they argue that if $v$ is an aribtrary vertex with connected component of size $k$, then there is probability $O(p^k)=O(n^{-k})$ that $G(n,p)$ has more than $k-1$ edges in this connected component. Here $p=c/n$.
I'm having trouble seeing why this is the case. The connected component has $\binom{k}{2} - (k-1)$ other possible edges to consider. We want the probability that at least one of them is on. Isn't this $1-(1-p)^{\binom{k}{2} - (k-1)}$? If so, how is this expression $O(p^k)$? Furthermore, what is changing in the big-oh notiation here? Is $k \to \infty$?
Edit:
This comes up in the proof the following fact: $$P[T_c^{Po}=k] = \lim_{n \to \infty} P[|C(v)|=k].$$
Here $T_c^{Po}$ is the progeny size of the branching process with Poisson distribution with mean $c$.
Let $S$ be a set of $k$ vertices. There are $\binom{\binom k2}{k}$ ways to choose a set $E$ of $k$ edges. For any particular such set $E$, the probability is $p^k$ that all $k$ edges are present.
If we want to know whether $S$ spans at least $k$ edges, this is the union of all $\binom{\binom k2}{k}$ of these events. The events are not independent and in fact there is lots over overlap. However, we can write $$ \Pr[S \text{ spans at least $k$ edges}] \le \binom{\binom k2}{k} p^k $$ by the union bound: the probability would be maximized if the events were disjoint.
This probability is $O(p^k)$ if we are considering $n \to \infty$, $p = \frac cn$, and $k$ constant. In this case, $\binom{\binom k2}{k}$ is just another (really big) constant, so it can be ignored.