2014 AMC 10A Problem 8

Note that all of these answer choices are of the form $$\dfrac{n!(n + 1)!}{2} = \dfrac{(n!)^2(n + 1)}{2}.$$ We have that $(n!)^2$ is square, so we need $\dfrac{n + 1}{2}$ to be square as well. This means that $n + 1$ must be twice a perfect square. The only choice we have is $n + 1 = 18,$ which gives us $n = 17.$ Thus, D is the correct answer.