expanding a factorial by simplifying with $(k-1)!$

Question

I'm wandering how $\frac{(k+2)!}{(k-1)!} = \frac{(k+2)(k+1)(k)(k-1)!}{(k-1)!}$. I'm confused about how $(k-1)!$ can be inserted into the numerator here. I know that this eventually works out to $k^3+3k+2k$, but why would $(k-1)!$ be justified here from the numerator $(k+2)!$?

Thank you

Answer 1

Because by definition of factorial we have $$ (k+2)!=(k+2)(k+1)(k)\underbrace{(k-1)(k-2)\cdots3\cdot 2\cdot 1}_{(k-1)!} $$

Answer 2

You may note that $(k+2)!$ is the product of integers from $1$ to $(k+2)$ that is $$ (k+2)!= 1\times 2 \times 3 ... \times (k-1)(k)(k+1)(k+2) =(k-1)!(k)(k+1)(k+2)$$