The image you've uploaded includes three expressions to simplify. Let's tackle each one step by step:

a) $\frac{n!}{(n-1)!} \cdot n$

To simplify this:

1. First, recognize that $\frac{n!}{(n-1)!}$ can be written as $n$, because $n! = n \cdot (n-1)!$.
2. This results in: $\frac{n!}{(n-1)!} \cdot n = n \cdot n = n^2$ So the simplified expression is $n^2$.

b) $\frac{(n+2)!}{(n-1)!} \cdot (n+2) \cdot (n+1) \cdot n$

To simplify:

1. Start with $\frac{(n+2)!}{(n-1)!}$. Recall that $(n+2)! = (n+2)(n+1)n(n-1)!$.
2. Thus, $\frac{(n+2)!}{(n-1)!} = (n+2)(n+1)n$.
3. Multiply by the other terms $(n+2)(n+1)n$ and $n$: $(n+2)(n+1)n \cdot (n+2)(n+1)n = (n+2)^2 (n+1)^2 n^2$ Thus, the simplified expression is: $(n+2)^2 (n+1)^2 n^2$

c) $\frac{(n+1)! + n!}{2n!} \cdot \frac{n+2}{2}$

To simplify:

1. Recognize that $(n+1)! = (n+1) \cdot n!$, so $(n+1)! + n! = (n+1)n! + n! = n!(n+2)$.
2. This makes the first part: $\frac{(n+1)! + n!}{2n!} = \frac{n!(n+2)}{2n!} = \frac{n+2}{2}$
3. Multiply by $\frac{n+2}{2}$: $\frac{n+2}{2} \cdot \frac{n+2}{2} = \frac{(n+2)^2}{4}$

Thus, the simplified expression is: $\frac{(n+2)^2}{4}$

Would you like further explanations or details on any part of the solution? Here are some related questions:

1. How do you simplify expressions involving factorials more generally?
2. What are the properties of factorials that help simplify expressions?
3. How can you handle higher-order terms like $(n+2)!$ in similar problems?
4. What are some common mistakes when simplifying expressions with factorials?
5. How does understanding factorial relationships help in simplifying more complex algebraic expressions?

Tip: Always look for opportunities to cancel out terms, especially when dealing with factorials, as they often contain repeating elements that can simplify your work.