Prove an upper bound for the binomials This is (supposed to be) an upper bound on the binomial coefficient:

$$ \binom{n}{k} \le \frac{n^n}{k^k(n-k)^{n-k}}$$

If we prove it by induction for all integers $0 \le k \le n/2$, we can easily show that it generalizes for $k \le n$, because $\binom{n}{k} = \binom{n}{n-k}$.

How can we prove it by induction? I get to:

$$ \binom{n}{k} \le \frac{n-k+1}{k}\frac{n^n}{(k-1)^{k-1}(n-k+1)^{n-k+1}} $$

And this is where I get stuck.
$$ n^n=(k+(n-k))^n=\sum_{j=0}^n{n\choose j}k^j(n-k)^{n-j}\ge {n\choose k}k^k(n-k)^{n-k}.$$
I personally thinks that there is an pitfall in Daniel Fischer's solution, but I can not comment directly because lack of reputation.

Daniel used Bernoulli inequality to prove $1-\frac{n-1}{{(n+1)}^2}\le{(1-\frac1{(n+1)^2})}^{n-1}$, but the actually form of Bernoulli inequality is $1 + rn \le {(1+r)}^n$, where r is a constant, but here $r = -\frac1{(n+1)^2}$ is a function of n.

So, when can't apply Bernoulli inequality directly, but follow the induction technique.
${(1-\frac1{(n+1)^2})}^{n} \ge  {(1-\frac1{(n+1)^2})} \times {(1-\frac{n-1}{(n+1)^2})}$
   $ = 1- \frac{n}{(n+1)^2} + \frac{n-1}{(n+1)^4}$, but what to prove is $ {(1-\frac1{(n+2)^2})}^{n} \ge 1-\frac{n}{{(n+2)}^2}$, it is trivial that $ {(1-\frac1{(n+2)^2})}^{n} \ge {(1-\frac1{(n+1)^2})}^{n} $, but it is easy verify that $ 1- \frac{n}{(n+1)^2} + \frac{n-1}{(n+1)^4} \not \ge 1-\frac{n}{{(n+2)}^2}$, so we  can't conclude that $ {(1-\frac1{(n+2)^2})}^{n} \ge 1-\frac{n}{{(n+2)}^2}$.