Riddles, brain puzzles and mathematical problems

[Froge]

Prove that n must be odd for (1+1/2)(1+1/3)(1+1/4)...(1+1/n) to be an integer.

Denote X as the set of natural numbers greater than or equal to 2.

P(n):

(1+1/2)(1+1/3)...(1+1/n) ∈ Z <-> n = 2k+1, k ∈ Z, k >= 1 for n ∈ X. Proof by induction.

P(1): (1 + 1/2) = 3/2, not an integer, but: (1 + 1/2)(1 + 1/3)
= 1 + 1/2 + 1/3 + 1/6
= 2 is an integer.

Suppose P(k) is true for k ∈ X. If k is odd:

(1+1/2)....(1+1/k) = A ∈ Z.

Consider now k+1, which is even:
(1+1/2)....(1+1/k)(1+1/(k+1)) = A(1 + 1/(k+1))
Given 1 + 1/(k+1) cannot be an integer for k ∈ X, A(1+1/(k+1)) is thus not an integer.

Consider now k+2, which is odd:
(1+1/2)....(1+1/k)(1+1/(k+1))(1+1/(k+2)) = A(1 + 1/(k+1))(1+1/(k+2))
=A(1 + 1/(k+1) + 1/(k+2) + 1/(k+1)(k+2))
=A(1 + (k+2 + k+1 + 1)/((k+1)(k+2)))
=A(1 + (2k + 4)/((k+1)(k+2)))
=A(1 + 2(k+2)/((k+1)(k+2)))
=A(1+ 2/(k+1))
=A((k+3)(k+1))

....I'm stuck, but I suspect a similar procedure follows for if k is even, and then by principle of mathematical induction, P(n) is true.