# Identity for generalized hypergeometric function

I want to show $$(z \frac{d}{dz} + a_j){}_p F_q(a_1,...,a_j,...a_p;b_1,...,b_q;z) = a_j {}_p F_q(a_1,...,a_j+1,...a_p;b_1,...,b_q;z),$$ as is found here.

My problem is, that I'm stuck at this step: $$(z \frac{d}{dz} + a_j){}_p F_q(a_1,...,a_p;b_1,...,b_q;z) =\\ z \frac{a_1...a_p}{b_1...b_q}F_q(a_1+1,...,a_p+1;b_1+1,...,b_q+1;z) + a_j{}_p F_q(a_1,...,a_p;b_1,...,b_q;z)$$

For simplicity we will show this for $$a_1$$ and denote $$\mathbf a=\{a_2,\dots,a_p\}$$, $$(\mathbf a)_k=\prod_{l=2}^p(a_l)_k$$, $$\mathbf b=\{a_1,\dots,a_q\}$$, and $$(\mathbf b)_k=\prod_{l=1}^q(b_l)_k$$. Then, \begin{align} (z \tfrac{d}{dz} + a_1){_pF_q}(a_1,\mathbf a;\mathbf b;z) &=\sum_{k=0}^\infty\frac{(a_1)_k(\mathbf a)_k}{(\mathbf b)_k\,k!}(z \tfrac{d}{dz}z^k + a_1z^k)\\ &=\sum_{k=0}^\infty\frac{(a_1)_k(\mathbf a)_k}{(\mathbf b)_k\,k!}(zkz^{k-1}+ a_1z^k)\\ &=\sum_{k=0}^\infty\frac{(a_1)_k(\mathbf a)_k}{(\mathbf b)_k\,k!}(k+ a_1)z^k. \end{align} But $$(k+ a_1)(a_1)_k =a_1\frac{(a_1+k)\Gamma(a_1+k)}{a_1\Gamma(a_1)} =a_1\frac{\Gamma(a_1+1+k)}{\Gamma(a_1+1)} =a_1(a_1+1)_k,$$ and so $$(z \tfrac{d}{dz} + a_1){_pF_q}(a_1,\mathbf a;\mathbf b;z) =a_1\sum_{k=0}^\infty\frac{(a_1+1)_k(\mathbf a)_k}{(\mathbf b)_k\,k!}z^k=a_1{_pF_q}(a_1+1,\mathbf a;\mathbf b;z),$$ which is the desired result.