lower bound in sorting algorithm

In geeksforgeeks.com there is an interesting fact about the lower bound of the sorting algorithm . the link is :---- http://www.geeksforgeeks.org/lower-bound-on-comparison-based-sorting-algorithms/

please explain this part which I felt a tricky part ---

 n!  <= 2^x

 Taking Log on both sides.
      \log_2n!  <= x

 Since \log_2n!  = \Theta(nLogn),  we can say
      x = \Omega(nLog_2n)