Solution:
Let $N = 2017^{2025!}$. Let $p$ be a prime dividing $2025!$ other than $2017$. Let $p^{k}$ be the largest power of $p$ dividing $2025!$. Clearly, $\phi (p^{k}) = (p - 1)p^{k - 1}$ divides $2025!$ and $\gcd (2017, p^{k}) = 1$, so by Euler's Totient Theorem,
$$N \equiv 1 \pmod{p^{k}}.$$
Repeating for all such primes $p$, we obtain
$$N \equiv 1 \pmod{2025! / 2017}.$$
Therefore, $\frac{2025!}{2017} \mid N - 1$, so $r = \frac{2025!}{2017} s$ for some $0 \leq s < 2017$. Also, since $N \equiv 0$ (mod $2017$), we have $r = \frac{2025!}{2017} s \equiv -1$ (mod $2017$).
By Wilson's,
$$\frac{2025!}{2017} = 2016! (2018)(2019)\ldots (2025) \equiv -8! \equiv 20 \pmod{2017}.$$
Therefore, $s$ is negative the inverse of $20$ (mod $2017$), which is $1311$. Our answer is
$$\frac{r}{2025!} = \frac{(2025! / 2017)(1311)}{2025!} = \frac{1311}{2017}.$$