Why/How Does $(N-1)! =N!/N$

Question

I'm studying for a test and in one of the solutions to a problem, Stirling's Approximation is used and allows the equation

$$ \frac{(q+N-1)!} {q! \cdot (N-1)!} $$

to become

$$ \frac{1}{q!} \cdot \frac{N}{N!} \cdot \frac{(q+N)!}{q+N} $$

I don't understand how the last two terms came to be. Is it a factorial identity or is there underlying math that hasn't been shown in the solution. Any help would be great.

Answer 1

$$\require{cancel}n!=\overbrace{1\times2\times3\times\dots\times(n-1)\times n}^n$$

$$\frac{n!}n=\frac{\overbrace{\color{green}{1\times2\times3\times\dots\times(n-1)}\times\color{red}{\cancel n}}^n}{\color{red}{\cancel n}}$$

$$(n-1)!=\overbrace{\color{green}{1\times2\times3\times\dots\times(n-1)}}^{n-1}$$

We also have:

$$\frac{n!}{p!}=\frac{\overbrace{\color{red}{\cancel{1\times2\times3\times\dots\times p}}\times\color{green}{(p+1)\times\dots\times(n-1)\times n}}^n}{\color{red}{\cancel{1\times2\times3\times\dots\times p}}}$$

$$=\color{green}{(p+1)\times\dots\times(n-1)\times n}$$

Answer 2

The factorial can resursively defined as $0! = 1$ and $$ n! = n\cdot(n-1)!$$ for $n\geq 1$.

An example is $4! = 4\cdot 3! = ... = 4\cdot 3\cdot 2\cdot 1.$ Hence by isolating the $(n-1)!$ term we get $$(n-1)! = \frac{n!}{n}.$$

Answer 3

$$ \frac{(q+N-1)!} {q! \cdot (N-1)!}=\frac{1}{q!}\cdot\frac{1}{(N-1)!}\cdot(q+N-1)!= \\ =\frac{1}{q!}\cdot\frac{N}{N\cdot(N-1)!}\cdot\frac{(q+N-1)!\cdot(q+N)}{(q+N)}= \frac{1}{q!}\cdot\frac{N}{N!}\cdot\frac{(q+N)!}{(q+N)} $$