003/POTW 297 Secondary school/ High school students

https://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-297-jan-17th-2018-a-23325.html#post104312

Problem:
if $k_1,k_2,\cdots k_{15}$ are the roots of the equation $x^{15} - 2x^{14} + 3 x^{13} + \cdots + 15x-16= 0$ evaluate $(1+k_1)(1+k_2)\cdots(1+k_{15})$

Because $k_1,k_2,\cdots k_{15}$ are the root

Hence $(x-k_1)(x-k_2)\cdots(x-k_{15}) = x^{15} - 2x^{14} + 3 x^{13} + \cdots + 15x-16$ as RHS is a degree 15 polynomial and coefficient is 1

putting $x= -1 $ we get
$(-1-k_1)(-1-k_2)\cdots(-1-k_{15}) = -1 - 2 -3 - \cdots - 15-16= -136$
or
$(-1)^{15}(1+k_1)(1+k_2),\cdots(-1+k_{15}) = -136$
or $(1+k_1)(1+k_2)\cdots(-1+k_{15}) = 136$