2020 AMC 12B Problem 6

Factor $(n + 1)!$ from the numerator: $$\begin{gathered} (n + 2)! - (n + 1)! \\ {}= (n + 1)!\,\big[(n + 2) - 1\big] \\ {}= (n + 1)!\,(n + 1). \end{gathered}$$ Dividing by $n!$ leaves $\dfrac{(n + 1)!\,(n + 1)}{n!}$ $= (n + 1)(n + 1)$ $= (n + 1)^2,$ which is always a perfect square. Thus, the correct answer is D.