$\prod_{I=0}^{24} \frac{n- i - 1}{n - i}$ have closed form?

Question

I am evaluating the closed form of

$\prod_{I=0}^{24} \frac{n- i - 1}{n - i}$ where $n > 25 \in N$

I can see that I cam get closed form of this if I do multiplication 25 times. Yet, I would like to know whether it is possible to get the form by other ways.

Answer 1

Observe \begin{align} \frac{n-1}{n}\frac{n-2}{n-1}\cdots \frac{n-24}{n-23}\frac{n-25}{n-24} = \frac{n-25}{n} \end{align}

or

\begin{align} \prod^{24}_{i=0} \frac{n-i-1}{n-i} =&\ \exp\left(\sum^{24}_{i=0}\left[\log(n-i-1)-\log(n-i) \right] \right)\\ =&\ \exp\left(\log(n-25)-\log(n) \right) = \frac{n-25}{n} \end{align} by Telescoping sum.

Answer 2

$$\begin{align} \prod_{i=0}^{24}{n-i-1\over n-i} &=\left(n-1\over n \right)\left(n-2\over n-1 \right)\left(n-3\over n-2 \right)\cdots\left(n-24\over n-23 \right)\left(n-25\over n-24 \right)\\ &={1\over n}\left(n-1\over n-1 \right)\left(n-2\over n-2 \right)\cdots\left(n-24\over n-24 \right)(n-25)\\ &={n-25\over n}\\ &=1-{25\over n} \end{align}$$