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

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