<s>[INST] Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$ Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$$
How to start the proof by induction? Is there any way to show this?
 [/INST] How to start a proof by induction?  
What a straight line!  
You start at the beginning.
...
Then you show that the middle flows, and conclude that the end is inevitable.  
But seriously... You start with $n = 2$.  
Prove that $(1 - 1/4) = (2+1)/2*2$.  
That's the initial or base step.
Then you do the induction step.  You prove that if $(1-1/4)....(1-\frac{1}{k^2}) = \frac{k+1}{2*k}$ for some $k$, then it follows that $(1-1/4) .....(1-\frac{1}{(k+1)^2}) = \frac{k + 2}{2(k+1)}$
And then you are done.  It follows that if it is true for $n =2$, and you've proved that if it is true for $n = k$ then it must be true for $n = k + 1$, you have proven it is true for all $n \ge 2$.
The heart is showing that  if $(1-1/4)....(1-1/k^2 = (k+1)/2*k$ for some $k$, then it follows that $(1-1/4) .....(1-\frac{1}{(k+1)^2}) = \frac{k + 2}{2(k+1)}$
For that note $(1-1/4) .....(1 - \frac{1}{k^2})(1-\frac{1}{(k+1)^2}) = [![![![![(1-1/4) .....(1 - \frac{1}{k^2})]!]!]!]!](1-\frac{1}{(k+1)^2}) = [\frac{k+1}{2*k}][1-\frac{1}{(k+1)^2})]$
Can you prove that that equals $\frac{k + 2}{2(k+1)}$?
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