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

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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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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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