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Compute the product from n = 2 to infinity of [1 - 1/n^2]

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Please provide multiple ways to solve the following:

sin(pi z)/(pi z) = Product from n = 1 to infinity of [1 - z^2/n^2] (1)

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This solution we compute the product using both the infinite product representation of sin(pi z) and using elementary, high school level, and methods.

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We have:

sin(pi z)/(pi z) = Product from n = 1 to infinity of [1 - z^2/n^2] (1)

To compute the product from n = 2 to infinity of (1 - 1/n^2), we need to take out the factor for n = 1 in the product in (1), because this is zero at z = 1:

sin(pi z)/(pi z) = Product from n = 1 to infinity of [1 - z^2/n^2] =

(1 - z^2) Product from n = 2 to infinity of [1 - z^2/n^2] ------------>

Product from n = 2 to infinity of [1 - z^2/n^2] = sin(pi z)/[pi z (1 - z^2)]

We can then take the limit of z to 1 on both ...

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