Confused between comparison of asymptotic notation e^n vs 2^2n

Question

If we compare e^n and 2^2n, then since e > 2 that means 2^2n = O(e^n). But, I was just wondering if we can assume e > 2 here, and also can we ignore the constants from power as well?

Answer 1

Assuming here you mean 2²ⁿ and not 2²n (writing 2^2n is a bit ambiguous).

You are right that e > 2 but consider the rules of exponentiation. We have that:

2²ⁿ = (2²)ⁿ = 4ⁿ.

Now, 4 > e, so in fact eⁿ = O(2²ⁿ).

And the reverse is not true: 2²ⁿ ≠ O(eⁿ). The base of the power does matter in big-O notation as it creates different complexity classes. This constant you cannot ignore.