algorithm - O notation proof for exponents and power -

i trying prove 4^n not in order of o(2^n).

is valid method ?
4^n >= c*2^n => 4^n/2^n >= c => 2^n >= c

i got lost here...

well, method concrete. should proceed in same direction. currently, don't have better option.

4^n = ((2^2)^n) = (2^2n) = (2^n) * (2^n) > 2^n values of n>0. 

as, (2^n) * (2^n) > o(2^n) .

this because (2^n) * (2^n) > c * 2^n. therefore,there doesn't exist constant value greater 2^n!

hence, 4^n != o(2^n) 4^n > 2^n values of n>0.