Find the closed form of this product

The product of 1 - 1/(k^2) from k=2 to N.

I've searched online for product formulas; I found closed form summation formulas, but nothing for product. I've tried a number of things, but nothing works out. How do I find this product?

Note that Π(k = 2 to N) (1 - 1/k^2)
= Π(k = 2 to N) (k^2 - 1)/k^2
= Π(k = 2 to N) [(k - 1)(k + 1)]/k^2
= Π(k = 2 to N) (k - 1)/k * Π(k = 2 to N) (k + 1)/k.

Now, this becomes easy.
Π(k = 2 to N) (k - 1)/k = (1/2) * (2/3) * (3/4) * ... * (N - 1)/N = 1/N
Π(k = 2 to N) (k + 1)/k = (3/2) * (4/3) * (5/4) * ... * (N + 1)/N = (N+1)/2.

Therefore, Π(k = 2 to N) (1 - 1/k^2) = (1/N) * (N+1)/2 = (N + 1)/(2N).
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I hope this helps!

I wrote out of few of the terms and found the pattern:

product of 1 - 1/(k^2) from k=2 to N = (N+1) / (2N).

I'm working on the "How" part, I've just given you the "what".

EDIT: To figure this out, I first rewrote it as product (k^2-1)/k^2, or of (k+1)*(k-1)/k^2. As you take successive products starting with k, you see a pattern emerge:

(k-1)/k * (k+2)/(k+1) (product of first two terms, in this case N=3)

Next product:

(k-1)/k * (k+3)/(k+2) (product of first three terms, in this case N=4).

For any general N then, we have

(k-1)/k * (k+(N-1))/(k+(N-2)). Note: You can prove this rigorously using induction.

Now plug in k=2, simplify to get the solution.