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Establish a formula for the product (1-1/2)(1-1/3)…(1 – 1/n)

Claim:

    \[ \left( 1 - \frac{1}{2} \right) \left(1 - \frac{1}{3} \right) \cdots \left( 1 - \frac{1}{n} \right) = \frac{1}{n}. \]

Proof. If n = 2 on the left we have (1-1/2) = 1/2 and on the right we have (1/n) = 1/2. Thus, the formula is true for the case n=2.
Assume then that the formula is true for some k = n \in \mathbb{Z}_{\geq 2}. So,

    \begin{align*}  && \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3} \right) \cdots \left(1 - \frac{1}{k} \right) &= \frac{1}{k} & (\text{Ind. Hyp.}) \\ \implies && \left( 1 - \frac{1}{2} \right) \left(1 - \frac{1}{3} \right) \cdots \left(1 - \frac{1}{k+1}\right) &= \left(\frac{1}{k} \right) \left(1- \frac{1}{k+1} \right) & (\text{Multiplying by } 1 - \frac{1}{k+1} )\\  &&&= \frac{1}{k} - \frac{1}{k+1} \\  &&& = \frac{k+1-1}{k(k+1)} \\  &&& = \frac{1}{k+1}. \end{align*}

Thus, if the formula is true for k then it is true for k+1. Since we have established that it is true for n =2, we have that is true for all n \in \mathbb{Z}_{\geq 2}. \qquad \blacksquare

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