The powers will be very large, so some lateral thinking is needed.
In the overall factorisation, each natural number, n, up to 22, will occur in f= (22-n+1) factorials.
e.g. 2 will occur in 21 factorials, 3 in 20, and 22 will occur in 1.
If f is even, we are done, because the overall of repeats is even, giving an even power. In this puzzle, this includes every odd n, including odd primes.
If n is itself a square, then all of its repeats will also be squares. Hence we can ignore 3,4,5,7,9,11,13,15,16,17,19, and 21. Then we count the number of prime repeats in the remaining numbers,n.
e.g. 2 will occur in 2,6,8,10,14,18,22. (In 12 and 20, 2 occurs to an even power)
This is 7 occasions, so 2 will occur to an overall odd power.
3 and 5 occur an even number of times so are even powers.
7 occurs once (14) so is odd.
11 occurs once (22) so is odd.
So we desire to even up 2,7,11.
An obvious candidate is 11!=2^8×3^4×5^2×7×11
This deals with 7 and 11 but not 2.
Happily, 2! = 2 so we can remove that.
This should leave a perfect square (whatever it is) as desired.
So the two factorials are 2! and 11!
Possible Solution
