Prove that ${n \choose k} \ge (\frac{n}{k})^k$ for any given $n \ge k$ using induction I need to prove that  ${n \choose k} \ge (\frac{n}{k})^k$ for any given $n \ge k$ using induction. 
I started working on this but got stuck with the inductive step, namely I want to prove that 
$${n \choose k} \ge \left(\frac{n}{k}\right)^k \Longrightarrow {n \choose k+1} \ge \left(\frac{n}{k+1}\right)^{k+1}$$ 
This is how I started: 
$$\left( \frac{n}{k+1}\right)^{k+1} = \left(\frac{n}{k+1} \right) \left(\frac{n}{k+1}\right)^k \le \left(\frac{n}{k+1} \right) \left(\frac{n}{k} \right)^k \\ \le \underbrace{\left( \frac{n}{k+1}\right) {n \choose k}}_\text{hypothesis}$$
Right now, I don't see any reasonable way to continue with this. Any hints will be most appreciated. Thank you in advance for your answers.
 A: First, observe that the inequality doesn't make sense if $k=0$, so we assume $1\leq k\leq n$ throughout.
The formula ${n\choose k}=\frac{n(n-1)\dots(n-k+1)}{k!}$ can be rewritten as
$${n\choose k}=\prod_{i=0}^{k-1}\frac{n-i}{k-i}$$
so it suffices to see that for each $i$ we have $\frac{n-i}{k-i}\geq\frac{n}{k}$, which is immediate.
You can disguise this argument as an induction on $k$ (the statement being obvious for $k=1$ and any $n$) by rewriting the above formula as
$${n\choose k}=\frac{n}{k}{n-1\choose k-1},$$
applying the induction hypothsesis to ${n-1\choose k-1}$, and noting as before that $\frac{n-1}{k-1}\geq\frac{n}{k}$.  
A: HINT
I think you should apply the induction on $n$, notably


*

*Base case $k=n\implies {k \choose k} \ge \left(\frac{k}{k}\right)^k$

*Induction step assuming true ${n \choose k} \ge \left(\frac{n}{k}\right)^k$ prove that ${n+1 \choose k} \ge \left(\frac{n+1}{k}\right)^k$

A: Following @gimusi hint, 
$\binom{n+1}{k} = \frac{n+1}{n+1-k} \binom{n}{k} \ge  \frac{n+1}{n+1-k} \left(\frac{n}{k}\right)^k = \left(\frac{n+1}{k}\right)^k \left(\frac{n}{n+1}\right)^k \frac{n+1}{n+1-k} = \left(\frac{n+1}{k}\right)^k \frac{1}{\left(1+\frac{1}{n}\right)^k} \frac{1}{\left(1 - \frac{k}{n+1}\right)}$
So if we can show that $\frac{1}{\left(1+\frac{1}{n}\right)^k} \frac{1}{\left(1 - \frac{k}{n+1}\right)} \ge 1$, then we are done.  
To prove:$\left(1+\frac{1}{n}\right)^k \left(1 - \frac{k}{n+1}\right) \le 1$
Even to a first degree, the above $= \left(1 + \frac{k}{n}\right)
 \left(1 - \frac{k}{n+1}\right) = 1 + \frac{k}{n(n+1)} - \frac{k^2}{n(n+1)}$
The case for $k = n+1$ is trivially correct. Assume $k = n$. 
So even then 
$1 + \frac{n}{n(n+1)} - \frac{n^2}{n(n+1)} = 1 + \frac{1}{n+1}- \frac{n}{n+1} = \frac{2}{n+1}$ and we know $0 < \frac{2}{n+1} \lt 1$
