$$ \sum\limits_{k=0}^{n} {{n\choose k} (-1)^k \frac{\theta}{\theta + k}} = \prod\limits_{k=1}^{n}{\frac{k}{\theta + k}}, \mbox{ for all } \theta > 0 \mbox{ and all } n \in \mathbb{N}$$
Could someone please show how to prove this binomial identity evaluating the left side of the equation as the evaluation of a hypergeometric function $_2F_1(-n,\theta ;\theta +1|1)$ and then applying the Chu-Vandermonde formula to get the right side?
Starting from the definition of the hypergeometric function from Wikipedia $$_2F_1(a,b;c;z) = \sum_{m\ge 0} \frac{(a)_m (b)_m}{(c)_m} \frac{z^m}{m!}$$ with $(q)_n$ the rising Pochhammer symbol, we see that $$_2F_1(-n,\theta;\theta+1; 1) = \sum_{m\ge 0} \frac{(-n)_m (\theta)_m}{(\theta+1)_m} \frac{1}{m!} = \sum_{m=0}^n \frac{(-n)_m (\theta)_m}{(\theta+1)_m} \frac{1}{m!}$$ because when $m>n$ there is a zero term in $(-n)_m.$ Continuing $$ \sum_{m= 0}^n (-1)^m \frac{n!}{(n-m)!} \frac{(\theta)_m}{(\theta+1)_m} \frac{1}{m!} = \sum_{m= 0}^n (-1)^m \binom{n}{m} \frac{\theta}{\theta+m}.$$ This is the left side.
For the right side, according to Wikipedia and Mathworld we have that $$_2F_1(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.$$ This gives $$_2F_1(-n,\theta;\theta+1; 1) = \frac{\Gamma(\theta+1)\Gamma(n+1)}{\Gamma(\theta+1+n)\Gamma(1)} = n! \prod_{k=1}^n \frac{1}{\theta + k} = \prod_{k=1}^n \frac{k}{\theta + k}.$$
Addendum. A proof by induction of this identity can be found at the following MSE Post. There is another proof by Mellin transforms at this MSE Post.
