Prove this for any $k>0$

Question

Prove that $k!>(\frac{k}{e})^{k}$.

It is known that $e^{k}>(1+k)$. So if we multiply $k!$ on both sides, we get $k!e^{k}>(k+1)!$. Also $k^k>k!$. Now how to proceed ?

Answer 1

For some reason I cannot comment so I'll share my thought here: Stirling's approximation

Answer 2

Work by induction.

• For $k=1$ it is true as $e >1$
• Assume it is true for $k$. Then $(k+1)! = (k+1) k! > k k!> k \frac{k^k}{e^k} = e \frac{k^{k+1}}{e^{k+1}} > \frac{k^{k+1}}{e^{k+1}}$ because $e>1$.

Answer 3

I take it that $k$ is a positive integer. Then the question is equivalent to

$e^k>\frac{k^k}{k!}$

But $e^k=\displaystyle \sum_{n=0}^\infty \frac{k^n}{n!}$ and one of the summands (all of them are positive) is itself $\frac{k^k}{k!}$