f(n) = O(g(n)) is true... how ? apply log continuously
LHS :-
after apply 1st LOG :- $ lg (2^{2^{2^{n}}}) = 2^{2^{n}} . lg 2 $
after apply 2nd LOG :- $ lg(2^{2^{n}} . lg 2 ) = lg(2^{2^{n}}) + lg (lg2)=2^{n}. ( lg 2) + lg ( lg (2))$ ( consider only dominating terms why because constants can not have any effect )
after apply 3rd LOG :- $ lg({2^{n}} . lg 2 ) = lg({2^{n}}) + lg (lg2)= {n}. ( lg 2) + lg ( lg (2)) $
RHS :-
after apply 1st LOG :- $ lg (n^{n^{n^{n}}}) = n^{n^{n}} . lg n $
after apply 2nd LOG :- $ lg(n^{n^{n}} . lg n ) = lg(n^{n^{n}}) + lg (lgn)=n^{n}. ( lg n) + lg ( lg (n))$ ( consider only dominating terms why because lower terms can not have any effect )
after apply 3rd LOG :- $ lg({n^{n}} . lg n ) = lg({n^{n}}) + lg (lgn)= {n}. ( lg n) + lg ( lg (n)) $
now compare LHS and RHS ===> RHS have lg n term extra ===> f(n) = O(g(n))
