I want to show that ${n \choose k}$ $p^{k(k-1)/2}$ $=$ $\Theta$$(n^{k}p^{k(k-1)/2})$ (for $n$$\rightarrow$$\infty$) where $p$$\in$$[0,1]$ and $k$ is a positive integer. I was able to show that ${n \choose k}$$p^{k(k-1)/2}$$=$$O$$(n^{k}p^{k(k-1)/2})$ but I am having difficulty showing $(n^{k}p^{k(k-1)/2})$$=$$O$${n \choose k}$$p^{k(k-1)/2}$.
This has come up in a probability course when trying to find a threshold function for $G(n,p)$ to contain a $k$-clique. If there are more suitable tags for this problem then please add them.
Because we are looking at the asymptotic behaviour when $n$ tends toward infinity, we don't care about the term $p^{k(k-1)/2}$. Thus we only need to show that $\begin{pmatrix} n\\k \end{pmatrix} = \Theta(n^k)$.
You already proved that ${n \choose k} = O\begin{pmatrix}n^{k} \end{pmatrix}$. Let us prove the other way.
For this, we shall prove that there exist $c > 0, n_0 \in \mathbb{N}$ so that $n^k \leq c \begin{pmatrix} n\\k \end{pmatrix},\ \forall n \geq n_0$.
Let $n_0=2k$. Then, $\forall n \geq n_0$ $$\begin{pmatrix} n\\k \end{pmatrix} = \frac{\prod_{i=0}^{k-1} (n-i)}{k!} \geq \frac{\prod_{i=0}^{k-1} n/2}{k!} = \frac{n^k}{2^k k!} = \frac{1}{c}n^k,$$ where $c = 2^k k!$ is independant of $n$.
Thus, we proved that $\begin{pmatrix} n\\k \end{pmatrix} = \Theta(n^k)$.