If induction is involved, I'm assuming you mean for all natural numbers n. Let's start with the base case of n = 1.
If n = 1, then 2^n = 2^1 = 2, and
n = 1, so LHS > RHS and the inequality holds true for n = 1.
For our induction hypothesis, we assume 2^k > k for some k > 1.
(We want to prove that 2^(k + 1) > (k + 1) )
2^(k + 1) = (2^k)(2^1)
which, by our induction hypothesis, is
= k + k
And since k > 1
> k + 1
2^(k + 1) > k + 1
Confusing? Here's the entire chain all over again.
2^(k + 1) = 2*(2^k) > 2*(k) = k + k > k + 1
2^(k + 1) > (k + 1)
Which means the inequality holds true for n = k + 1.
Thus by the Principal of Mathematical Induction,
2^n > n for all natural numbers n.