Show $(n_{i}+g_{i}-1)!\simeq(g_{i}-1)! g_{i}^{n_{i}}$

Question

I can' t comprehend why the the following is true: $$n_{i}<<g_{i}$$ $$(n_{i}+g_{i}-1)!\simeq(g_{i}-1)! g_{i}^{n_{i}}$$

Answer 1

$$ (n+g-1)!=(n-1+g)!=(n-1+g)(n-2+g)\ldots(1+g)g(g-1)! $$ and using $n << g$ approximate $(n-1+g)$ etc. by $g$: $$ (n-1+g) \approx g\\ (n-2+g) \approx g\\ \vdots\\ (1+g) \approx g $$ so $$ (n+g-1)!=g\,g\,\ldots g\,g\,(g-1)! $$ where there are $n$ terms in $g$, giving your result.

For a specific example if $g=1000$ and $n=10$ the approximations being made are $$ 1009\approx 1000\\ 1008\approx 1000\\ \vdots \\ 1001 \approx 1000 $$