Prove that $$_2F_1(1,-k;1-k;-t) = 1 + \frac{tk}{1-k}~{_2F_1}(1,1-k;2-k,-t), \quad(1)$$ where $k \in (0, 1)$, $t > 0$, and $_2F_1(a,b;c;z)$ is the Gauss hypergeometric function given by $$_2F_1(a,b;c;z) \triangleq \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-xt)^{-a}\mathrm{d}x.$$
I tried using the change of variables, but the addition 1 on the right-hand side of (1) creates the problem.
The particular case of the hypergeometric function in the left hand side of (1) has $a=1$, $b=-k$ and $c=1-k$ which can be represented as \begin{equation} \begin{aligned} {}_2F_1(1, -k; 1-k; \color{red}{z}) &= \frac{\Gamma(1-k)}{\Gamma(-k)\Gamma(1)}\int_0^1 \frac{dx}{x^{k+1}(1-x\color{red}{z})}\\ &=\frac{\Gamma(1-k)}{\Gamma(-k)\Gamma(1)}\int_0^1 \frac{1-xz + xz}{x^{k}x(1-xz)}dx\\ &= \frac{\Gamma(1-k)}{\Gamma(-k)\Gamma(1)} \int_0^1\left[\frac{1}{x^{k+1}} + \frac{z}{x^{k}(1-xz)} \right]dx. \end{aligned} \end{equation} Given the recursive property of Gamma function is $\Gamma(1)=1$ and $\Gamma(x+1)=x\Gamma(x)$, the factor in front of the integral is simplified to $-k$.
You can obtain the exact value of the first term inside the integral. You also need to rewrite the second term inside the integral to represent it as a hypergeometric function. After doing those, just replace $z=-t$ to get your desirable proof.