For all integers $n \geq 9,$ the value of
\[\frac{(n+2)!-(n+1)!}{n!}\]is always which of the following?
$\textbf{(A) } \text{a multiple of 4} \qquad \textbf{(B) } \text{a multiple of 10} \qquad \textbf{(C) } \text{a prime number} \qquad \textbf{(D) } \text{a perfect square} \qquad \textbf{(E) } \text{a perfect cube}$
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 $\boxed{\textbf{(D) } \text{a perfect square}}.$