2023 AMC 10B Problem 15

Group the product as $(2!\,3!)(4!\,5!)\cdots(14!\,15!)\cdot 16!.$ Since $(2k)!(2k+1)!$ $= \bigl((2k)!\bigr)^2(2k+1),$ each pair is a perfect square times an odd number. Those odd numbers $3, 5, 7, 9, 11, 13, 15$ multiply to $3^4 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13,$ with squarefree part $7 \cdot 11 \cdot 13.$ And $16! = 2^{15} 3^6 5^3 7^2 \cdot 11 \cdot 13$ has squarefree part $2 \cdot 5 \cdot 11 \cdot 13.$ Multiply the two: the squarefree part of the whole thing is $2 \cdot 5 \cdot 7.$ That’s the smallest $m,$ namely $2 \cdot 5 \cdot 7 = 70.$ Thus, C is the correct answer.