Simple concept, complicated equation

As this is a product, it will equal zero when at least of of the factors is zero. Therefore, we set the term inside the product equal to zero.

$\frac{(x-k)^{k+1}}{(x+k)^{k-1}}=0$ $(x-k)^{k+1}=0$ $x-k=0$ $x=k$

Where $k$ varies among the integers from 1 to 63. To find the sum of the roots, we sum all k's.

$\sum_{k=1}^{63} k$

Using the formula for the sum of the first n integers, we have

$\sum_{k=1}^{63} k = \dfrac{63\cdot 64}{2}$

Evaluating, we get $\boxed {2016}$