smc

We first expand the expression: $$\frac{(n+2)!-(n+1)!}{n!}=\frac{(n+2)(n+1)n!-(n+1)n!}{n!}.$$ We can now divide out a common factor of $n!$ from each term of the numerator: $$(n+2)(n+1)-(n+1).$$ Factoring out $(n+1),$ we get $$[(n+2)-1](n+1)=(n+1{)}^2,$$ which proves that the answer is $\boxed{\text{a perfect square}}.$