Prove the following statement by mathematical induction. For every integer n 0, 2n <(n + 2)!. Proof (by mathematical induction): Let P(n) be the inequality 2n < (n + 2)!. We will show that P(n) is true for every integer n 0. (a) Show that P(0) is true. (For each answer, enter a mathematical expression.) Before simplifying, the left-hand side of P(0) is . Before simplifying, the right-hand side of P(0) is . The fact that the statement is true can be deduced from that fact that 20 = 1. (b) Show that for each integer k 0, if P(k) is true, then P(k + 1) is true. (For each answer, enter a mathematical expression.) (i) Let k be any integer with k 0, and suppose that P(k) is true. In other words, suppose that . [This is P(k), the inductive hypothesis.] (ii) We must show that P(k + 1) is true. P(k + 1) is the inequality . (iii) Information about P(k + 1) can be deduced from the following steps. Identify the reason for each step. 2k < (k + 2)! by the induction hypothesis by the induction base by basic algebra 2 < k + 3 k + 3 < 2 2k 2 < (k + 2)! 2 by the induction hypothesis by the induction base by basic algebra 2 < k + 3 k + 3 < 2 2k + 1 < (k + 2)! 2 by the induction hypothesis by the induction base by basic algebra 2 < k + 3 k + 3 < 2 2k + 1 < (k + 2)! (k + 3) by the induction hypothesis by the induction base by basic algebra 2 < k + 3 k + 3 < 2 2k + 1 < (k + 3)! by the induction hypothesis by the induction base by basic algebra 2 < k + 3