Approximation of combination $\binom{n}{k}k!$

Question

Let $G$ be a graph with $k$ vertices and $e$ edges and let $X$ be the number of copies of $G$ in $G(n,p)$. The expectation of $X$ is $$ {n \choose k}\frac{k!}{|aut(G)|} p^e = \Theta \left( n^kp^e\right) $$

I am wondering how we can obtain $\Theta \left( n^kp^e\right)$. I can see that ${n \choose k}\frac{k!}{|aut(G)|} p^e\leq \frac{n^k}{k!}k!p^e=n^kp^e$, but I could not find any lower bound for it!

Answer 1

Hint

$$A=\binom{n}{k}\,k!=\frac {n!}{(n-k)!}$$ Take logarithms of both sides, use Stirling approximation twice and continue with Taylor series for "large" values of $n$.

When done, use Taylor again since $A=e^{\log(A)}$