Prove by induction the product of n odd integers is odd for n >1. Recall a number k is odd if there exists an integer m such that k= 2m+ 1.

Step-by-step explanation: Induction: Suppose that for n odd numbers is true that their product is odd. Lets find for n+1 numbers: The pruduct of n+1 odd numbers can be written as: (2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1)*(2kₙ₊₁ + 1) where the numbers "k" can be any integer number. Because of the hipotesis, we know that: (2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1) is an odd number, so we can write this as: (2k + 1) Then we have: (2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1)*(2kₙ₊₁ + 1) = (2k + 1)*(2kₙ₊₁ + 1) now we have to solve this product: (2k + 1)*(2kₙ₊₁ + 1) = 2k*2kₙ₊₁ + 2k + 2kₙ₊₁ + 1 = 2*(2k*kₙ₊₁ + k + kₙ₊₁) + 1 Because k and kₙ₊₁ are integers, the number (2k*kₙ₊₁ + k + kₙ₊₁) is also an integer, that we can write as L. So we have that: (2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1)*(2kₙ₊₁ + 1) = 2L + 1 So the product of n + 1 odd integers is also an odd number.  Now you can test that this is true for n = 2 (the same procedure that I did in the product of (2k + 1)*(2kₙ₊₁ + 1) ) and there you have your demonstration by induction.