Question

Prove that n! = Ω(n^100)

Answer 1

n! = Ω(n^100)

First "Ω" means ">=" in terms of asymptotic notations

so it says n! >= n^100  (this terminology should only be used at the

 time of solving mcqs)

Now taking log on both side we get

n(logn) >= 100(logn) {as log(n! = n logn)}

so now we by reducing logn from both side we get
n>= 100(constant).{which is true}
Hence Proved