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...

$\binom{n}{k}=\binom{n}{k-1}\frac{n-k+1}{k}$. And this fraction is larger than $1$ if $k<\frac{n}{2}$
