= Solution
For the base case $n=2$, the product is $1-1/4=3/4=(2+1)/(2\cdot2)$. Assume the identity holds for some $n\geq2$. Multiplying by the next factor gives
$$
\begin{aligned}
\prod_{i=2}^{n+1}\left(1-\frac1{i^2}\right)
&=\frac{n+1}{2n}\left(1-\frac1{(n+1)^2}\right)\\
&=\frac{n+1}{2n}\frac{n(n+2)}{(n+1)^2}
=\frac{n+2}{2(n+1)}.
\end{aligned}
$$
This is the desired expression with $n$ replaced by $n+1$, so <mathematical induction> proves the identity for every $n\geq2$. The factorization $1-i^{-2}=(i-1)(i+1)/i^2$ also explains the cancellation as a <telescoping product>.
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