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**Deveno** basically a strong inductive proof will run as follows:

base case: handled by ProveIt in post #2.

assume that for 1 < k < n odd numbers, their product is odd.

suppose we have n odd numbers {a_{1},...,a_{n}}. without loss of generality, assume the last number is a_{n} = 2t + 1, for some natural number t.

since a_{1}a_{2}...a_{n-1} is a (n-1)-fold product of odd numbers, and n-1 < n,

by our inductive hypothesis, this number is odd, so a_{1}a_{2}...a_{n-1} = 2u + 1, for some natural number u.

therefore a_{1}a_{2}....a_{n} = (a_{1}a_{2}...a_{n-1})(a_{n}) = (2u + 1)(2t + 1),

which is odd by our base case.