How many positive integer cubes are divisors of $1!\cdot 2! \cdot 3!\cdot 4! \cdot 5!\cdot 6!\cdot 7!\cdot 8!$? What I have tried:
$$N= 1!\cdot  2! \cdot 3!\cdot  4! \cdot 5!\cdot 6!\cdot  7!\cdot 8!\\
= 1^8\cdot 2^7\cdot 3^6\cdot 4^5\cdot 5^4\cdot 6^3\cdot 7^2\cdot 8^1 \\
= 2^{23}\cdot 3^9\cdot 5^4 \cdot 7^2$$

You're off to a good start.
Hint 1: Let $n^3$ be a positive integers cube. Suppose $n$ has the prime factorization $n = p_1^{a_1} p_2^{a_2} \cdots p_n^{a_n}$. What is the prime factorization of $n^3$? In general, what can you say about the exponents of the prime factors of a perfect cube?
Hint 2: Let's simplify the problem a bit and count the number of positive integer cubes that divide $3^9 5^4$. If $n^3$ divides $3^9 5^4$, then it can only have $3$ and $5$ as prime factors. By Hint 1, the possible exponents for $3$ are $0$, $3$, $6$ and $9$. Similarly, the possible exponents for $5$ are $0$ and $3$. So there are $4 \cdot 2 = 8$ total positive integer cubes that divide $3^9 \cdot 5^4$. Can you generalize this to your number?
