{
   \begin{eqnarray*}
      \frac{(2n)!}{{(n!)}^2}  & < & 2^{2n}\\
      {2n \choose n}          & < & 2^{2n}\\
      {2n \choose n}          & < & 2^{2n} = {2n \choose 0} + {2n \choose 1} +
         \ldots + {2n \choose 2n-1} + {2n \choose 2n}.\ \checkmark
   \end{eqnarray*}
   }