Prove $\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} = \prod_{k=1}^n \frac{k}{x+k}$ and more

$$\sum_{k=0}^n \dbinom{n}k (-y)^{x+k-1} = (-y)^{x-1} (1-y)^n$$ Hence, $$\int_0^1\sum_{k=0}^n \dbinom{n}k (-y)^{x+k-1}dy = \int_0^1(-y)^{x-1} (1-y)^n dy$$ $$\sum_{k=0}^n \dbinom{n}k (-1)^{x+k-1}\int_0^1y^{x+k-1}dy = \sum_{k=0}^n \dbinom{n}k (-1)^{x+k-1} \dfrac1{x+k}$$ $$\int_0^1(-y)^{x-1} (1-y)^n dy = (-1)^{x-1} \beta(x,n+1)$$ Hence, $$\sum_{k=0}^n \dbinom{n}k (-1)^{k} \dfrac1{x+k} = \beta(x,n+1)$$