Here, I want to show that the marginal distribution of $x$ given by $Z(x) = (n, x) B(x + 1, n - x + 1)$ is uniform on the interval of discrete support $[0, ..., n]$ where $(n, x)$ is the binomial coefficient where $n$ is the number of trials.
Suppose $n = 10$:
$\displaystyle Z(x = 0) = (10, 0) \left[\frac{\Gamma (1) \Gamma (n+1)}{\Gamma (n + 2)}\right] = 1 \left[\frac{1 x 3628800}{3.99168e7}\right] = 0.0909090..., Z(x = 1) = (10, 1) \left[\frac{\Gamma (2) \Gamma (n+1)}{\Gamma (n)}\right] = 10 \left[\frac{1 x 3628800}{3.99168e7}\right] = 0.0909090, ..., Z(x = 10) = (10, 10) \left[\frac{\Gamma (11) \Gamma (1)}{\Gamma (n + 2)}\right] = 0.0909090$
How do I show algebraically that the above is true for any $n$?
The Gamma function is $\Gamma(k)=(k-1)!$ whenever $k\in \{1,2,...\}$. This immediately yields $$\frac{n!}{x!(n-x)!}\frac{x!(n-x)!}{(n+1)!}=\frac{n!}{(n+1)!}=\frac{1}{n+1}$$