A Sheer Multiplication Of Numbers

Relevant wiki: Factorials Problem Solving - Intermediate

Consider the factorial : $1!\times 2!\times \ldots \times 119!\times 120!$

If we group the factorials into pairs, then we can factor out a $(n !)^2$ from each pair $n!\times (n+1)!$

Then, what is left is $(2\times 4\times \ldots 120)k^2$ , where $k^2= (1!)^2\times \ldots (119!)^2$ .

So, now we need to just look at $2\times 4\times \ldots 120 = 2^{60}(60!)$ , so the entire expression is $k^2\times 2^{60} \times 60!= (2^{30}k)^2 \times 60!$ .

So, we see that if we take out a $60!$ , then the rest of the expression would form a perfect square. So, our desired answer is $60!$