Is there a combinatorial way to see the link between the beta and gamma functions?
Here's a combinatorial argument for $a!\, b! = \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(a+b+1)!}{(a+i+1)}$, which is just a slight rewrite of the identity you want to show.
Suppose you have $a$ red balls numbered $1$ through $a$, $b$ blue balls numbered $1$ through $b$, and one black ball.
Question: How many permutations of the balls have all the red balls first, then the black ball, and then the blue balls?
Answer 1: $a! \,b!$. There are $a!$ ways to choose the red balls to go in the first $a$ slots, $b!$ ways to choose the blue balls to go in the last $b$ slots, and $1$ way for the black ball to go in slot $a+1$.
Answer 2: Let $A$ be the set of all permutations in which the black ball appears after all the red balls (irrespective of where the blue balls go). Let $B_i$ be the subset of $A$ such that the black ball appears after blue ball $i$. Then the number of permutations we're after is also given by $|A| - \left|\bigcup_{i=1}^b B_i\right|$. Since the probability that the black ball appears last of any particular $a+i+1$ balls is $\frac{1}{a+i+1}$, and there are $(a+b+1)!$ total ways to arrange the balls, by the principle of inclusion-exclusion we get $$\frac{(a+b+1)!}{a+1} - \sum_{i=1}^{b}\binom{b}{i}(-1)^{i+1}\frac{(a+b+1)!}{(a+i+1)} = \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(a+b+1)!}{(a+i+1)}.$$
Using partial fractions, we have that $$ \frac{1}{(a+1)(a+2)\dots(a+b+1)}=\frac{A_1}{a+1}+\frac{A_2}{a+2}+\dots+\frac{A_{b+1}}{a+b+1}\tag{1} $$ Use the Heaviside Method; multiply $(1)$ by $(a+k)$ and set $a=-k$ to solve $(1)$ for $A_k$: $$ A_k=\frac{(-1)^{k-1}}{(k-1)!(b-k+1)!}=\frac{(-1)^{k-1}}{b!}\binom{b}{k-1}\tag{2} $$ Plugging $(2)$ into $(1)$, yields $$ \frac{a!}{(a+b+1)!}=\sum_{k=1}^{b+1}\frac{(-1)^{k-1}}{b!}\binom{b}{k-1}\frac{1}{a+k}\tag{3} $$ Multiplying $(3)$ by $b!$ and reindexing, gives us $$ \frac{a!b!}{(a+b+1)!}=\sum_{k=0}^{b}(-1)^k\binom{b}{k}\frac{1}{a+k+1}\tag{4} $$ and $(4)$ is your identity.
Update: Starting from the basic binomial identity $$ (1-x)^b=\sum_{k=0}^b(-1)^k\binom{b}{k}x^k\tag{5} $$ multiply both sides of $(5)$ by $x^a$ and integrate from $0$ to $1$: $$ B(a+1,b+1)=\sum_{k=0}^b(-1)^k\binom{b}{k}\frac{1}{a+k+1}\tag{6} $$