Question 1020545
{{{(1-1/2)(2-2/3)(3-3/4)}}}...{{{(14-14/15)(15-15/16)}}}
Notice that the kth factor of this expression also has k as a factor,e.g., the factor 3-3/4 has 3 as a factor,and the 14th factor 14-14/15 has 14 as a factor.
If we factor all the k factors from each one of those factors, the expression becomes

{{{1*2*3*4}}}...{{{13*14*15(1-1/2)(1-1/3)(1-1/4)}}}...{{{(1-1/15)(1-1/16)}}}
This expression is the same as 
{{{15!(1/2)(2/3)(3/4)}}}...{{{(14/15)(15/16)}}}

The product of the fifteen fractional terms would just be equal to 1/16 by direct cancellation of common terms.
Hence the entire expression is equal to 

{{{15!(1/16)}}}, 

and so the value of n is {{{highlight(15)}}}.