Prove that if n ∈ ℕ where n > 1, then n + 1 is odd.

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SUMMARY

For any natural number n greater than 1, n! (factorial of n) is always even due to the presence of 2 as a factor. Consequently, when 1 is added to an even number, the result is odd. Therefore, it is established that n! + 1 is odd for all n ∈ ℕ where n > 1.

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Dustinsfl
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Prove that if n ∈ ℕ where n > 1, then n! + 1 is odd.
n = k + 1, k = 1,...,∞
Now substitute n!+1 = (k+1)! + 1 = ?

I am not sure what to do next or if that is on the right path.
 
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If n > 1 then n! is an even number since it has 2 as one of its factors, so n! is even for n > 1. Conclude.
 

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