To prove the statement using mathematical induction:
**Base Case:**
When n=1, the statement becomes 1 < 2^1, which is true.
**Inductive Hypothesis:**
Assume that for some integer k, k < 2^k.
**Inductive Step:**
We need to show that k+1 < 2^(k+1).
We know that k < 2^k by our inductive hypothesis. Also, we know that 2^k < 2^k+1 because multiplying a number by 2 is equivalent to shifting its binary representation to the left by one position.
Therefore,
k < 2^k < 2^k+1
Adding 1 to both sides of the inequality chain, we get:
k+1 < 2^k+1
Multiplying both sides by 2, we get:
2(k+1) < 2^(k+1)
Since 2^(k+1) = 2*2^k, we get:
2(k+1) < 2*2^k
k+1 < 2^k
This completes the inductive step.
Therefore, by mathematical induction, the statement n < 2^n is true for all positive integers n.