My teamwork claimed that a good approximation for $n \choose k$ is $(\frac{n}{k}\cdot e)^k$. he said to me that in most cases $n \choose k$ is really closed to $(\frac{n}{k}\cdot e)^k$ (up to factor of 1.25).
What is the worst case approximation ratio between $n \choose k$ and $(\frac{n}{k}\cdot e)^k$? I know that it can be bounded by $e$ (since $ (\frac{n}{k})^k< {n \choose k} < (\frac{n}{k}\cdot e)^k $), my question is there is a better approximation ratio for this.