As a sort of lead in, consider the simpler situation S = 1!.2!.3!.4!.5!.6!.7!.8! = (1).(1.2).(1.2.3). ... .(1.2.3.4.5.6.7.8).
It's easier to see the product if it's written as,
1! = 1
2! = 1.2
3! = 1.2.3
4! = 1.2.3.4
5! = 1.2.3.4.5
6! = 1.2.3.4.5.6
7! = 1.2.3.4.5.6.7
8! = 1.2.3.4.5.6.7.8
S will equal the product of all of the numbers on the rhs of the equals signs.
Notice that all of the even numbers 2, 4, 6, 8 occur an odd number of times, while the odd numbers 3, 5, 7 each occur an even number of times, (we can ignore the 1's).
\(\displaystyle S = 2^{7}.3^{6}.4^{5}.5^{4}.6^{3}.7^{2}.8^{1}, \\ = (2^{7}.4^{5}.6^{3}.8^{1}).(3^{6},5^{4}.7^{2}) ,\\ = (2^{6}.4^{4}.6^{2}).(2.4.6.8).(3^{6}.5^{4}.7^{2}), \\ =(2^{6}.4^{4}.6^{2}).2^{4}.(1.2.3.4).(3^{6}. 5^{4}.7^{2}), \\ = (2^{6}.4^{4}.6^{2}).2^{4}.4!.(3^{6}.5^{4}7^{2}).\)
Apart from the 4! in the middle everything else is a perfect square, so if we divide S by 4! the result will be a perfect square.
The same routine can be used for S = 1!.2!.3!. ... 100! .
The factorial in the middle will be 50!.
This only works if the final factorial is a multiple of 4. Try this for S = 1!.2!.3!.4!.5!.6! for example, and it doesn't work because the power of 2 that appears in the middle will be 3, so not a perfect square.
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