Product of Sequence of 1 minus Reciprocal of Squares/Proof 1

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
 * $\displaystyle \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$

It is first noted that $n = 0$ is excluded because in that case $\dfrac {n + 1} {2 n}$ is undefined.

$\map P 1$ is the other edge case:

Thus $\map P 1$ is seen to hold.

Basis for the Induction
$\map P 2$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\displaystyle \prod_{j \mathop = 2}^k \paren {1 - \dfrac 1 {j^2} } = \dfrac {k + 1} {2 k}$

from which it is to be shown that:
 * $\displaystyle \prod_{j \mathop = 2}^{k + 1} \paren {1 - \dfrac 1 {j^2} } = \dfrac {k + 2} {2 \paren {k + 1} }$

Induction Step
This is the induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 1}: \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$