Not So Tedious Multiplication

Relevant wiki: Telescoping Series - Product

We can cancel out the numbers from the numerator of a fraction and a denominator from another fraction in pairs:

$\begin{aligned} P & = \frac 12 \cdot \frac 23 \cdot \frac 34 \cdots \frac {3441}{3442} \cdot \frac {3442}{3443} = \frac 1{\cancel{2}} \cdot \frac {\cancel{2}}{\cancel{3}} \cdot \frac {\cancel{3}}{\cancel{4}} \cdots \frac {\cancel{3441}}{\cancel{3442}} \cdot \frac {\cancel{3442}}{3443} = \boxed{\dfrac 1{3443}} \end{aligned}$

consider the first three terms

$\frac{1}{2}$ $*$ $\frac{2}{3}$ $*$ $\frac{3}{4}$ $=$ $\frac{1}{4}$

2 and 2 cancels out and 3 and 3 cancels out

so the answer to the problem is $\frac{1}{3443}$