$\frac{N}{n}$; probability

Question

Here is an interesting identity:

$$(n\leq N)~\dfrac{N}{n}=1+\dfrac{N-n}{N-1}+\dfrac{(N-n)(N-n-1)}{(N-1)(N-2)}+\cdots+ \dfrac{(N-n)(N-n-1)\cdots 2\cdot 1}{(N-1)(N-2)\cdots (n+1)n}. $$

I failed to prove that; but I was given a hint that we could do this by probability method.

Then how to apply probability method to the proof? Any help would be appreciated!

Edit: The nice answer below seems to not have applied probability theory. How to do it by probability theory?

Answer 1

Writing the sum as: $$1+\dfrac{N-n}{N-1}+\dfrac{(N-n)(N-n-1)}{(N-1)(N-2)}+\cdots+ \dfrac{(N-n)(N-n-1)\cdots 2\cdot 1}{(N-1)(N-2)\cdots (n+1)n}\\ =1+\dfrac{N-n}{N-1}\left(1+\dfrac{N-n-1}{N-2}\left(1+\dfrac{N-n-2}{N-3}\cdots\left(1+\dfrac{2}{n+1}\left(1+\dfrac{1}{n}\right)\right)\right)\right)\\ \cdots =1+\dfrac{N-n}{N-1}\left(1+\dfrac{N-n-1}{n}\right)=1+\dfrac{N-n}{n}=\dfrac{N}{n}.$$ one observe that it "telescopes" from the end, leading to the final identity.