A Highly powered question

$1!2!3!4! ... 29!30! = (1)(1.2)(1.2.3)(1.2.3.4) .... (1.2.3 ...29)(1.2.3. ... 29.30)$

$= 1^{30}.2^{29}.3^{28}.4^{27} ... 29^2.30^1$

$= 2^{(29+27+25+23+...+1)+(27+23+19+15+...+3)+(23+15+7)+(15)}$

$$ $$ $$ $$ $$ $\times 3^{(28+25+22+...+1)+(22+13+4)+(4)} \times 5^{(26+21+16+...+1)+(6)}$

$$ $$ $$ $$ $$ $\times 7^{(24+17+10+3)} \times 11^{(20+9)} \times 13^{(18+5)} \times 17^{14} \times 19^{12} \times 23^8 \times 29^2$

$= 2^{225+105+45+15} \times 3^{145+39+4} \times 5^{81+6} \times 7^{54} \times .....$ can be ignored as powers are $<30$

$= 2^{390}3^{188}5^{87}7^{54} ...$ $$ $$ $$ $= (2^{30})^{13}.(3^{30})^6.(5^{30})^2.(7^{30})^1...$ primes with powers $<30$

$$

So the number is perfect $30^{th}$ power of each term in expansion of

$(2^0+2^1+2^2+...2^{13})(3^0+3^1+3^2+...3^6)(5^0+5^1+5^2)(7^0+7^1)$

$\implies 14.7.3.2 = \boxed{588}$ perfect $30^{th}$ power divisors