Nested Hypergeometric series Is it possible to express the following series as a hypergeometric function:
$$\sum_{n=0}^\infty (a)_n \sum_{j_1+j_2+\cdots+j_k=n} \frac{1}{(b)_{j_1} (b)_{j_2}\cdots (b)_{j_k}} z^n $$
where $(a)_n, (b)_n$ are Pochhammer symbols. 
Intuitively, if the inner sum can be expressed as a Pochhammer symbol, we obtain a hypergeometric series.
Any ideas, suggestions and clues are much appreciated.

$$\sum_{j_1+\ldots+j_k=n}\frac{z^{n}}{(b)_{j_1}\cdots (b)_{j_k}}\tag{1} $$
is the coefficient of $x^n$ in the product:
$$\left(\sum_{m\geq 0}\frac{z^m x^m}{(b)_m}\right)^k = \left(\int_{0}^{1}x z e^{txz}(1-t)^{b-1}\,dt\right)^k=x^k z^k\left(\int_{0}^{1}t^{b-1} e^{(1-t)xz}\,dt\right)^k\tag{2}$$
I wrote $(b)_m=\frac{\Gamma(b+m)}{\Gamma(b)}=\frac{\Gamma(m)}{B(b,m)}$, exploited the integral definition of the Beta function and switched $\sum$ and $\int$. So we have:
$$\sum_{n\geq 0}(a)_n \sum_{j_1+\ldots+j_k=n}\frac{z^{n}}{(b)_{j_1}\cdots (b)_{j_k}}= \frac{z^k}{\Gamma(a)}\sum_{n\geq 0}\Gamma(a+n)\cdot [x^{n-k}]\left(\int_{0}^{1}t^{b-1} e^{(1-t)xz}\,dt\right)^k\tag{3}$$
and the RHS of $(3)$ should be easy to rearrange in a hypergeometric format.
