The hypergeometric differential equation is given by $$(z(z\frac{d}{dz}+a)(z+\frac{d}{dz}+b)-(z\frac{d}{dz})(z\frac{d}{dz}+c-1))y(z) = 0$$
Now I want to show that the following is a solution of it $$z^{1-c}F(a+1-c;b+1-c;2-c \ | \ z)$$
How can I show that this is a solution of the hypergeometric differential equation? I already showed that the differential equation can be written as $$z(1-z)y''+(c-z(a+b+1))y'-aby = 0$$
But I don't know how to continue at this stage, since we only proved it for easier forms of the hypergeometric series