My calculator don't have that much storage

We can rewrite this as:

$\displaystyle \prod_{k=1}^{2015} \frac{k(k+1)}{(k+2)(k+3)}$

$\displaystyle =\prod_{k=1}^{2015} \left[ \frac{k}{k+1} \cdot \frac{k+1}{k+2} \cdot \frac{k+1}{k+2} \cdot \frac{k+2}{k+3} \right]$

Solving the four telescopic products:

$\displaystyle =\frac{1}{2016} \cdot \frac{2}{2017} \cdot \frac{2}{2017} \cdot \frac{3}{2018}$

And simpliflying, we get

$\displaystyle =\frac{1}{2^4 \cdot 3 \cdot 7 \cdot 1009 \cdot 2017^2}$

And our answer is $2+3+7+1009+2017=\boxed{3038}$

This fraction could be simplified as

$\huge \frac {(2015!) \times (2016!)}{ \huge \frac {2017!}{2} \times \frac {2018!}{6} }$

By simplification, this becomes

$\huge \frac { 2^{2} \times 3 \times {\color{#D61F06}{2015! }} \times {\color{#D61F06}{2016}!} } {\huge (2017 \times 2016 \times {\color{#D61F06}{2015!}}) \times (2018 \times 2017 \times {\color{#D61F06}{2016!}}) }$

We easily cancel the numbers in red, then we factorize the numbers, which yield us

$\huge \frac {2^{2} \times 3} { 2017^{2} \times 1009 \times 2^{6} \times 3^{2} \times 7}$

Thus, cancelling the terms we get

$\large ( 2^{4} \times 3 \times 7 \times 1009 \times 2017^{2} )^{-1}$

Giving us

$\large 2 + 3 + 7 + 1009 + 2017 = {\color{#3D99F6} {3038}}$ .

The fraction can be factored as $\frac{k(k+1)}{(k+2)(k+3)}$ , so consider the expansion of the product from this form:

$\frac{(1)(2)}{(3)(4)}*\frac{(2)(3)}{(4)(5)}*\frac{(3)(4)}{(5)(6)}*...*\frac{(2014)(2015)}{(2016)(2017)}*\frac{(2015)(2016)}{(2017)(2018)}$

Note that the numerator of the 3rd fraction matches the denominator of the first fraction. This pattern would continue until the end, and allows us to reduce almost everything, leaving only the following:

$\frac{(1)(2)(2)(3)}{(2016)(2017)(2017)(2018)}$

By reducing the numerator with the 2016, we get:

$\frac{1}{(168)(2017^{2})(2018)}$

It is helpful to point out that 2017 and 1009 are primes, so the prime factorization of the denominator goes quickly into:

$\frac{1}{(2^{4})(3)(7)(1009)(2017^{2})}$

Thus, $2+3+7+1009+2017=\boxed{3038}$