I am trying to prove that the series expansion of product of two Bessel functions of the first kind around $x=0$ is given by: $$ J_{\nu}(x)J_{\mu}(x) = \sum_{r=0}^\infty \frac{(-1)^r \Gamma(\mu+\nu+2r+1)}{r! \Gamma(\nu+r+1)\Gamma(\mu+r+1)\Gamma(\mu+\nu+r+1)}\left(\frac{x}{2}\right)^{\mu+\nu+2r} \tag{1} $$ where $\nu, \mu\in \mathbb{R}$.
My approach is to multiply the series expansions of the individual Bessel functions and rearrange the terms in order to obtain (1). $$ J_{\nu}(x)J_{\mu}(x) = \left(\sum_{n=0}^\infty \frac{(-1)^n }{n! \Gamma(n+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2n}\right) \left(\sum_{m=0}^\infty \frac{(-1)^m }{m! \Gamma(m+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2m} \right) $$ With some effort, I managed to reduce the above equation to: $$ J_{\nu}(x)J_{\mu}(x) = \sum_{r=0}^\infty \frac{(-1)^r }{r! \Gamma(\nu+r+1)\Gamma(\mu+r+1)}c_r \left(\frac{x}{2}\right)^{\mu+\nu+2r} \tag{2} $$ where $c_r$ is defined as: $$c_r = \sum_{n=0}^r \binom{r}{n} \left\{\prod_{i=r-n+1}^r (\nu+i)\right\} \left\{\prod_{j=n+1}^r (\mu+j)\right\}$$
So, it turns out that to prove (1), we need to prove that: \begin{align*} c_r&= (\mu+\nu+2r)(\mu+\nu+2r-1)\cdots (\mu+\nu+r+1)\\ &= \frac{\Gamma(\mu+\nu+2r+1)}{\Gamma(\mu+\nu+r+1)} \tag{3} \end{align*} For small values of $r$, one can verify that (3) is indeed correct. One way to proceed would be to use induction on $r$. So far I have been unsuccessful with that. I would really grateful if someone could provide an elementary proof of equation (3). Maybe there's a combinatorial interpretation that I am unable to understand. Any references/suggestions related to this problem are also welcome.
First recall that $a^{\overline{b}}=a(a+1)\cdots (a+b-1)$ is the definition of the rising factorial. Notice, furthermore, that your identity can be rewritten as $$\sum _{n=0}^r\binom{r}{n}(\nu +r-n+1)^{\overline{n}}(\mu +n+1)^{\overline{r-n}}=(\mu +\nu+r+1)^{\overline{r}}.$$
To take out the $n's$ inside the monomials notice that $(a-n+1)^{\overline{n}}=(a)^{\underline{n}}$ (here $a^{\underline{n}}$ is called the falling factorial defined as $a(a-1)\cdots (a-n+1)$) and so $(\nu +r-n+1)^{\overline{n}}=(\nu+r)^{\underline{n}}$ and $(\mu +n+1)^{\overline{r-n}}=(\mu +r)^{\underline{r-n}}$ so your left hand side becomes $$\sum _{n=0}^r\binom{r}{n}(\nu +r)^{\underline{n}}(\mu +r)^{\underline{r-n}}=(\nu +\mu+2r)^{\underline{r}}=(\nu +\mu +2r-r+1)^{\overline{r}},$$ where the intermediate step is the binomial theorem for falling factorials.