Formula for Geometric Progression Can someone help me understand the idea behind constructing a formula for the following:

For $n\in\mathbb{N}$, $n\geq 2$, find and prove a formula for:
  $$\prod_{i=2}^n \left(1 - \frac{1}{i^2}\right).$$

Please, Please, I need to know HOW to do this not just how to solve this question. 
Confused about:
- Why does I start at 2 and not 1 like regular series, is it because it makes the solution zero?
- Why does $n \geq 2$ have to be true?
- Do you start with $i = 2$ for the first element?
Thanks any help is appreciated!
 A: You can start a summation or a product anywhere, not just at $1$. There is no law mandating that they must start at $1$.
In this case, if you were to allow $i=1$, then the first factor of the product would be
$$\left(1 - \frac{1}{1^2}\right) = 0$$
and so the entire product would be zero, which would make everything rather silly. So you start at $2$ instead (you could instead reindex and consider
$$\prod_{j=1}^{n-1}\left(1 - \frac{1}{(j+1)^2}\right)$$
but that just makes the expression look more complicated; better to change the indices instead).
Why do you have $n\geq 2$? Because if $n\lt 2$, then the product has no factors (since no value of $i$ can be at least $2$ and also less than or equal to some $n\lt 2$). The empty product is equal to $1$ by definition, but most people have trouble with that assertion, so why introduce it? Better to stick to $n\geq 2$ so that it is clear there are always factors.
Yes: you start by plugging in $i=2$, then $i=3$, then $i=4$, and so on until you get to $i=n$. For example, with $n=4$, you have
\begin{align*}
\prod_{i=2}^4\left(1 - \frac{1}{i^2}\right) &= \left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{4^2}\right)\\
&= \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right)\\
&= \left(\frac{3}{4}\right)\left(\frac{8}{9}\right)\left(\frac{15}{16}\right)\\
&= \frac{360}{576} = \frac{5}{8}.
\end{align*}
As to finding a formula, well, you might try a few values and see if you spot a pattern; then you can try using induction to prove it.
Added. Okay, you say you are stuck, you don't see a pattern. This is just 
Consider the factors you are multiplying: $1 - \frac{1}{i^2}$. Writing them as a single fraction, you have
$$1 - \frac{1}{i^2} = \frac{i^2 - 1}{i^2} = \frac{(i-1)(i+1)}{i^2}.$$
So, what are you doing at each step? Let's write them out explicitly:
$$\left(\frac{(1)(3)}{2^2}\right)\left(\frac{(2)(4)}{3^2}\right)\left(\frac{(3)(5)}{4^2}\right)\cdots\left(\frac{(n-1)(n+2)}{n^2}\right).$$
The denominator is easy: you can just rearrange it, throw in a couple of extra factors of $1$ (who cares if you multiply by $1$? Doesn't do anything) and write:
\begin{align*}
&(2)(2)(3)(3)(4)(4)\cdots(n)(n)\\
&=(2)(3)(4)\cdots(n)(2)(3)(4)\cdots(n)= \Bigl((1)(2)(3)(4)\cdots(n)\Bigr)\Bigl((1)(2)(3)(4)\cdots(n)\Bigr).
\end{align*}
Now, what do you call the product of all the positive integers from $1$ through $n$? 
Great! We now have a formula for the denominator.
What about the numerator? Again, reorder them putting first all the first factors, and next all the second factors; we have:
\begin{align*}
&(1)(3)(2)(4)(3)(5)(4)(6)\cdots (n-1)(n+1)\\
&= \Bigl( (1)(2)(3)(4)\cdots(n-1)\Bigr)\Bigl((3)(4)(5)\cdots(n+1)\Bigr).
\end{align*}
Hmmm... Looks like another set of those "products of all positive integers from $1$ up to something." Now, granted, the second factor isn't that, but just throw in a $(1)(2)$, and then divide by $2$ to keep it honest:
$$\Bigl((1)(2)(3)(4)\cdots(n-1)\Bigr)\Bigl((1)(2)(3)(4)(5)\cdots(n+1)\Bigr)\left(\frac{1}{2}\right).$$
Now, express it using the same kind of symbol as the denominator.
Now, step back and look at what you get. Is there some simplification/cancellation that you can make to get something simple? What?
A: HINT: $$\prod_{i=2}^n \left( 1 - \frac{1}{i^2}\right) = \prod_{i=2}^n \left(\frac{i^2 - 1}{i^2} \right)=\prod_{i=2}^n \frac{(i-1)(i+1)}{i^2} = \frac{\displaystyle\prod_{i=2}^n (i-1) \prod_{i=2}^n (i+1)}{\left(\displaystyle\prod_{i=2}^n i\right)^2}.$$
