Proof that $\frac{n}{N}\cdot\frac{n-1}{N-1}\cdot\frac{n-2}{N-2}\cdot\ldots\cdot \frac{1}{N-n+1} = \frac{1}{{N\choose n}}$

Question

Is there a sophisticate way to proof that:

$$\frac{n}{N}\cdot\frac{n-1}{N-1}\cdot\frac{n-2}{N-2}\cdot\ldots\cdot \frac{1}{N-n+1} = \frac{1}{{N\choose n}}$$

where ${N\choose n}$ denotes combinations.

When replacing $N$ and $n$ with values, both produce the same result.

Answer 1

Simplify the numerator and denominator.

$$\frac{n}{N}\cdot\frac{n-1}{N-1}\cdot\frac{n-2}{N-2}\cdot\ldots\cdot \frac{1}{N-n+1} = \frac{n!}{\frac{N!}{(N-n)!}}$$

Simplify by $n!$.

$$\frac{1}{\frac{N!}{n!(N-n)!}}$$

The denominator is in the form $n$ choose $r$.

$$\frac{1}{N \choose n}$$

Answer 2

This is a definition: $ {N \choose n} = \frac{N!}{n!(N-n)!} $

Expand all factorials and simplify the tail (N-n)!