You can also proof it "by story". Let "random number" mean a number picked from $[0,1]$ with uniform probability. Then the formula below can be interpreted as follows.
$$\int_0^1 p^k (1-p)^{n-k} \mathrm{d}p = \frac{k!(n-k)!}{(n+1)!}$$
The left-hand side is the probability of taking random $p$ and then drawing a sequence of $n$ numbers from which some $k$ numbers are smaller than $p$ and some $n-k$ are larger.
To understand the right-hand side, consider $n+1$ random numbers sorted, so that first $k$ are the smallest and last $n-k$ are the largest (with $(k+1)$-th being our $p$ from left-hand side interpretation); however, there are $(n+1)!$ permutations total, with $k!(n-k)!$ having the desired property (in a sorted sequence we disregard the order of first $k$ and last $n-k$), thus the right-hand side fraction denotes the same probability.