Factorial simplificaton involving negative 1

Question

What is the best way of simplifying

$$\dfrac{(a+b-1)!}{(b-1)!}$$

Ideally i want to get rid of the two $-1$ and the final solution should not containt the gamma function

Answer 1

$$a!~{a+b\choose a}\frac b{a+b}~=~a!~{a+b\choose b}\frac b{a+b}$$

Answer 2

Assuming that $a$ and $b$ are natural numbers: $$ \frac{(a+b-1)!}{(b-1)!}=b(1+b)(2+b)\cdot\ldots\cdot(a-1+b). $$