I love $2015$

$2015=31\times13\times5$

The greatest $n$ such that $2015^n|2015! \Rightarrow n=\lfloor\frac{2015}{31}\rfloor+\lfloor\frac{2015}{31^2}\rfloor=65+2=67$

$\therefore$ the greatest $n$ such that $2015^n|2015!^{2015}$ is $67 \times 2015=135005$

$\boxed{n=135005}$

The prime factorization of 2015 = 31 X 13 X 5. In prime factorization of (2015!)^2015, we have 13^334490, 5^10011530, and 31^135005. Thus, 135005 31's multiply with 135005 5's and then with 135005 13's to form 135005 2015's. What is left in the numerator is 320990 13's, 9876525 5's and 0 31's.
There won't be more 2015 in the numerator because there is no more than 135005 31's to multiply and form 2015.

Read the solution carefully and I am sure you would understand me.