I encountered an easy(if you remember your limit calculus and derivatives) of showing big-O in some cases.  In one of the videos, TM mentioned little-o, which is a stronger version of big-O.  Essentially, f = O(g) means g grows at least as fast as f, while f = o(g) means g grows much faster than f.  So if f = o(g), then clearly f = O(g).

How can we use this to prove big-O easily?

Well, the definition of f = o(g) is equivalent to lim(f/g) = 0.  So if you can show lim(f/g) = 0, then f = o(g) and f = O(g).  If you cannot show that lim(f/g) = 0, then you'll have to prove it another way.

Even if you cannot show lim(f/g) = 0, 

you can show that lim(f/g) = c , (where c is a constant) then f = O(g).

But in this case, f = O(g) but not f = o(g). because g is not much greater than f.