Binomial Coefficient Inequality, prove $\binom{n}{0} < \binom{n}{1} < \binom{n}{2}< ... <\binom{n}{\left \lfloor {\frac{n}{2}}\right \rfloor}$ I don't know how to prove this inequality
$$\binom{n}{0} < \binom{n}{1} < \binom{n}{2}< ... <\binom{n}{\left \lfloor {\frac{n}{2}}\right \rfloor}$$
Knowing that
$$
(n-2k)\binom{n}{k}=n \left [\binom{n-1}{k}-\binom{n-1}{k-1}\right]
$$
The exercise explicitly asks me to use the equality (that is easy to prove) to show that the inequality holds. 
Any suggestion? Maybe I'm missing a basic trick...

Applying the equality to $n+1$ and $k < \frac{n+1}{2}$, you get
$$
(n+1)\left(\binom{n}{k} - \binom{n}{k-1}\right) = (n+1-2k)\binom{n+1}{k} > 0
$$
so $\binom{n}{k} - \binom{n}{k-1} > 0$.
