a) You should multiply and divide the given product by factorial `(n-1)!` such that:

`((n-1)!*n*(n+1)*(n+2)*(n+3))/((n-1)!)`

You need to remember that `(n+3)! = 1*2*3*...*(n-1)*n*(n+1)(n+2)(n+3) = (n+2)!(n+3) = (n+1)!*(n+2)*(n+3) = n!*(n+1)*(n+2)*(n+3) = (n-1)!*n*(n+1)*(n+2)*(n+3) `

Hence, you may substitute `(n+3)! for ((n-1)!*n*(n+1)*(n+2)*(n+3))` such that:

`n*(n+1)*(n+2)*(n+3) = ((n+3)!)/((n-1)!) `

**Hence, using factorials, you may write the product `n*(n+1)*(n+2)*(n+3)` such that: `n*(n+1)*(n+2)*(n+3) = ((n+3)!)/((n-1)!) ` **

b) Reasoning by analogy yields:

`(n+6)(n+7)(n+8) = ((n+5)!(n+6)(n+7)(n+8))/((n+5)!)`

`(n+6)(n+7)(n+8) = ((n+8)!)/((n+5)!)`

**Hence, using factorials, you may write the product `(n+6)(n+7)(n+8)` such that: `(n+6)(n+7)(n+8) = ((n+8)!)/((n+5)!).` **

c) **You need to arrange the terms of product such that:**

`(n-1)!*n = n!`

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