Answer
$\frac{1}{k+1}$
Work Step by Step
General formula : $\pi_{i=2}^n(1-\frac{1}{i})=\frac{1}{n}$ for all integers $n\geq 2$
Proof (by mathematical induction) let the property p(n) by the equation
$\pi^n_{i=2}(1-\frac{1}{i})=\frac{1}{n}$
Show that P(2) is true:
The left-hand side of P(2) is
$\pi^n_{i=2}(1-\frac{1}{i})=\frac{1}{n}=1-\frac{1}{2}=\frac{1}{2}$
which equals the right-hand side.
Show that for all integers k ≥ 2, if P(k) is true then
P(k + 1) is also true:
Suppose that k is any integer with k ≥ 2 such that
$\pi^k_{i=2}(1-\frac{1}{i})=\frac{1}{k}$
We must show that
$\pi^{k+1}_{i=2}(1-\frac{1}{i})=\frac{1}{k+1}$
But by the laws of algebra and substitution from the inductive hypothesis, the left-hand side of P(k + 1) is
$\pi^{k+1}_{i=2}(1-\frac{1}{i})$
=$\pi^{k}_{i=2}(1-\frac{1}{i})(1-\frac{1}{k+1})$
=$(\frac{1}{k})(1-\frac{1}{k+1})=(\frac{1}{k})(\frac{(k+1)-1}{k+1})$
=$\frac{1}{k+1}$ which is the right-hand side of P(k + 1)