It is proved by mathematical induction that when there is always 2 ^ n > n ^ 3 for a large enough natural number n, the first value no should be the minimum when the first step inequality is established
Let's look at the second one first
N = k holds
Then 2 ^ (K + 1) = 2 ^ k * 2 > 2K & sup3;
It is obvious that 2K & sup3; > (K + 1) & sup3;
That is (the cube root of K * 2) & sup3; > (K + 1) & sup3;
The cube root of K * 2 is more than K + 1
k> 1 / (cube root of 2-1)
1 / (cube root of 2-1) is about 3.8
So the minimum n is 4