Saudi Arabian Mathematical Competitions

Consider
$a_{1} = 2^{2010},\ a_{2} = 3^{2010} - 1,\ a_{3} = 4^{2010} - 1,\ \ldots,\ a_{2010} = 2011^{2010} - 1$
and get
$$\begin{gathered} a_{1}(1!)^{2010} + a_{2}(2!)^{2010} + \ldots + a_{2010}(2010!)^{2010} \\ = (2!)^{2010} + (3!)^{2010} - (2!)^{2010} + (4!)^{2010} - (3!)^{2010} + \ldots + \\ + (2011!)^{2010} - (2010!)^{2010} = (2011!)^{2010} \end{gathered}$$
Moreover, it is clear that $a_{1} < a_{2} < a_{3} < \ldots < a_{2010}$.