TITLE: Calculate the product of these $98$ rational numbers
QUESTION [11 upvotes]: $$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) ... \left( 1 + \frac{1}{99} \right) $$
Of course I could do it factor for factor but there has got to be a more efficient way. Does anyone know it?

REPLY [3 votes]: If you don't like the way mathematicians immediately rewrite fractions as $3(+)\frac17$ to $\frac{22}7$ (which suggests replacing $1+\frac1n$ by $\frac{n+1}n$), you might like the following approach. Contemplate the sequence 
$$
  2,3,4,5,6,7,\ldots,100.
$$
The second term is $\frac12$ times larger than the first, then the third is $\frac13$ times larger than the second (so that it is $(1+\frac12)(1+\frac13)$ times as large as the first term), then the fourth term is $\frac14$ times larger again, etc. Now how many times as large is the final term as the first one?