Prove that for all the natural numbers n that 2^n > n
Base Case is easy
Then the inductive step you have 2^k > k as the inductive hypothesis
show that p[k+1] holds
2^(k+1) > k+1
on the left side 2^(k+1) = 2^k * 2
but idk what else to do
Prove that for all the natural numbers n that 2^n > n
Base Case is easy
Then the inductive step you have 2^k > k as the inductive hypothesis
show that p[k+1] holds
2^(k+1) > k+1
on the left side 2^(k+1) = 2^k * 2
but idk what else to do