This is problem 6.9 in Stein complex analysis.
Let $$F(\alpha,\beta,\gamma;z) = 1+\sum_{n=1}^\infty{\alpha(\alpha+1)\cdots(\alpha+n-1)\beta(\beta+1)\cdots(\beta+n-1)\over n!\gamma(\gamma+1)\cdots(\gamma+n-1)}z^n$$ be the hypergeometric series for $\alpha,\beta\in\Bbb C$ and $\gamma\neq 0,-1,-2,\cdots$. Note that the ratio test implies this series has the radius of convergence $1$.
For $\alpha>0,\beta>0,\gamma>\beta$ and $|z|<1$, show that $$F(\alpha,\beta,\gamma;z) = {\Gamma(\gamma)\over\Gamma(\beta)\Gamma(\gamma-\beta)}\int_0^1 t^{\beta-1}(1-t)^{\gamma-\beta-1}(1-zt)^{-\alpha}\ dt.$$ [Hint: To prove the integral identity, expand $(1-zt)^{-\alpha}$ as a power series.]
My attempt: Direct computation shows that the Taylor expansion of $(1-zt)^{-\alpha}$ at $z =0$ is $$(1-zt)^{-\alpha} = 1+\sum_{n=1}^\infty{\alpha(\alpha+1)\cdots(\alpha+n-1)t^n\over n!}z^n.$$ Since $\Gamma$ function has no zero, it suffices to show $${\Gamma(\beta)\Gamma(\gamma-\beta)\over\Gamma(\gamma)}F(\alpha,\beta,\gamma;z) =\int_0^1 t^{\beta-1}(1-t)^{\gamma-\beta-1}(1-zt)^{-\alpha}\ dt.\tag{$\dagger$}$$ Note that $${\Gamma(\beta)\Gamma(\gamma-\beta)\over\Gamma(\gamma)} = B(\beta,\gamma-\beta) = \int_0^1 t^{\beta-1}(1-t)^{\gamma-\beta-1}\ dt,$$ where $B$ is a beta function.
Then the LHS of $(\dagger)$ becomes $$\int_0^1 t^{\beta-1}(1-t)^{\gamma-\beta-1}\ dt+\sum_{n=1}^\infty{\alpha(\alpha+1)\cdots(\alpha+n-1)\beta(\beta+1)\cdots(\beta+n-1)\over n!\gamma(\gamma+1)\cdots(\gamma+n-1)}z^n\int_0^1 t^{\beta-1}(1-t)^{\gamma-\beta-1}\ dt$$
The RHS of $(\dagger)$ equals
$$\int_0^1 t^{\beta-1}(1-t)^{\gamma-\beta-1}\ dt+\sum_{n=1}^\infty{\alpha(\alpha+1)\cdots(\alpha+n-1)\over n!}z^n\int_0^1 t^{\beta+n-1}(1-t)^{\gamma-\beta-1}\ dt.$$
where the infinite sum can escape the integral by the uniform convergence for $|z|<1$.
To remove $n$ in the integrand, tried integration by parts with $t^{\beta+n-1}$ and $(1-t)^{\gamma-\beta-1}$ but $n$ on $t$ moves to $(1-t)$ so $n$ does not disappear. I'm stuck at this point. Please help.