Nested binomial sum

[Tolaso J Kos]

The following nested binomial sum was conjectured by akotronis at the greek site mathematica.gr

Prove or disprove the following identity:

$$\sum_{a_1=0}^{b_1}\sum_{a_2=0}^{b_2}\cdots\sum_{a_n=0}^{b_n}\frac{\binom{b_1}{a_1}\binom{b_2}{a_2}\cdots\binom{b_n}{a_n}}{\binom{b_1+b_2+\ldots+b_n}{a_1+a_2+\ldots+a_n}}=b_1+b_2+\cdots+b_n+1$$

[r9m]

We may begin with the beta function identity for non negative integer values of $a,b$,

$$\int_0^1 x^{b-a}(1-x)^a\,dx = \frac{1}{(b+1)\binom{b}{a}}$$

Hence, for non-negative integers $a',b'$, \begin{align*}\sum\limits_{a = 0}^{b}\frac{\binom{b}{a}}{\binom{b+b'}{a+a'}} &= (b+b'+1)\sum\limits_{a = 0}^{b}\binom{b}{a}\int_0^1 x^{b+b'-a-a'}(1-x)^{a+a'}\,dx\\&= (b+b'+1)\int_0^1 x^{b'-a'}(1-x)^{a'}\,dx\\&= \frac{b+b'+1}{(b'+1)\binom{b'}{a'}} \tag{$*$}\end{align*}

As a result we may compute the nested summation as,

\begin{align*}\sum\limits_{a_1 = 0}^{b_1}\cdots \sum\limits_{a_n = 0}^{b_n}\frac{\binom{b_1}{a_1}\cdots \binom{b_n}{a_n}}{\binom{b_1+\cdots+b_n}{a_1+\cdots+a_n}} &= \frac{b_1+\cdots + b_n +1}{b_1+\cdots+b_{n-1}+1}\sum\limits_{a_1 = 0}^{b_1}\cdots \sum\limits_{a_{n-1} = 0}^{b_{n-1}}\frac{\binom{b_1}{a_1}\cdots \binom{b_{n-1}}{a_{n-1}}}{\binom{b_1+\cdots+b_{n-1}}{a_1+\cdots+a_{n-1}}} & \text{ (evaluating the innermost sum with $(*)$)}\\&= \cdots \\&= \prod\limits_{j=0}^{n-2}\frac{b_1+\cdots+b_{n-j}+1}{b_1+\cdots+b_{n-j-1}+1}\sum\limits_{a_1=0}^{b_1}\frac{\binom{b_1}{a_1}}{\binom{b_1}{a_1}}\\&= b_1+\cdots+b_n+1\end{align*}