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:
$$N= 1!\cdot  2! \cdot 3!\cdot  4! \cdot 5!\cdot 6!\cdot  7!\cdot 8!\\
= 1^8\cdot 2^7\cdot 3^6\cdot 4^5\cdot 5^4\cdot 6^3\cdot 7^2\cdot 8^1 \\
= 2^{23}\cdot 3^9\cdot 5^4 \cdot 7^2$$

Now, for each prime factor, how many possible exponents are there?
For example, for the prime factor $2$, the exponent can be $0,3,6,9,12,15,18$ or $21$.
