Approximation of the falling factorial

Question

I found an approximation of the falling factorial for $k\ll N$: $$ (N)_k = N(N-1)\ldots(N-k+1)\approx N^k $$ Any idea how to get it?

Answer 1

Note that $(N)_{k} = \frac{N!}{(N-k)!}$. From there, we may use Stirling's Approximation:

$n!\approx \sqrt{2\pi n}n^{n}e^{-n}$:

Subtituting:

$(N)_{k} \approx \frac{\sqrt{2\pi N}N^{N}e^{-N}}{\sqrt{2\pi (N-k)}(N-k)^{N-k}e^{k-N}}$

$(N)_{k}\approx \frac{N^{N}}{(N-k)^{N-k}}\sqrt{\frac{N}{N-k}}e^{-k}$

$\boxed{(N)_{k}\approx \frac{N^{N+\frac{1}{2}}}{(N-k)^{N-k + \frac{1}{2}}}e^{-k}}$

Answer 2

Factoring (N)_k = N^k (1- 1/N)(1-2/N)...[1-(k-1)/N)]. Then since k<<N, k/N-> 0. Therefore, (N)_k ~ N^k (1)^k = N^k.