The answer is $4030$. If $d$ is an odd integer, the expression $5^{d}-1$ can be factored as $(5-1)\left(5^{d-1}+5^{d-2}+\ldots+1\right)$ where the second factor contains $d$ odd addends, that is, an odd number of odd addends, and is therefore odd. A number of the form $5^{d}-1$ with $d$ odd is therefore divisible by $4$, and by no higher power of $2$; since the expression in the problem is precisely a product of $2015$ factors of this form, the prime number $2$ appears in its factorization with an exponent of $2 \cdot 2015$, that is $4030$.