TITLE: Approximation of combination $ {n \choose k} = \Theta \left( n^k \right) $?
QUESTION [11 upvotes]: Is it a valid to say 
$$
{n \choose k} = \Theta \left( n^k \right)
$$
for any $n$ and $k$? If so, how to prove it? 
Note: $k$ is not a function of $n$. 
Note: Observed it here (page 5): 
 http://www.cs.berkeley.edu/~sinclair/cs271/n6.pdf

REPLY [3 votes]: We can even make a stronger statement that $${n \choose k} = O\left (\left( \frac{e \cdot n}{k} \right)^k \right)$$ For the proof, see this.