$$\text{exp} \left ( \frac{1 }{1-\frac{2}{\alpha }}T \times {}_2F_1\left (1,1-\frac{2}{\alpha }; 2-\frac{2}{\alpha },-T \right ) \right ) = \text{exp}\left ( \Gamma\left ( \frac{2}{\alpha} \right )\Gamma\left ( 1-\frac{2}{\alpha } \right ) T^{\frac{2}{\alpha }} k \right ) $$
This is what I have tried:
Using the relation in this question (I don't know the proof for this yet)
$$ \, _2{F}_1(a,b;c;z) = \frac{\Gamma (c)}{\Gamma (b) \Gamma (c-b)} \int_0^{\infty} t^{-b+c-1} \, (t+1)^{a-c} \, (t-z+1)^{-a} \, dt $$
Now we substitute $a=1, b= 1-\frac{2}{\alpha}; c= 2-\frac{2}{\alpha}, z=-T$.
This gives
$$ \, _2{F}_1(a,b;c;z) = \frac{\Gamma \big({2-\frac{2}{\alpha}}\big)}{\Gamma \big({1-\frac{2}{\alpha}}\big)\Gamma (1)} \int_0^{\infty} t^{2} \, (t+1)^{{-1+\frac{2}{\alpha}}} \, (t+T+1)^{-1} \, dt $$
On further simplification and multiplying with the remaining part of the LHS
$$ \frac{1 }{1-\frac{2}{\alpha }}T \times \frac{\Gamma \big({2-\frac{2}{\alpha}}\big)}{\Gamma \big({1-\frac{2}{\alpha}}\big)\Gamma (1)} \int_0^{\infty} t^{2} \, (t+1)^{{-1+\frac{2}{\alpha}}} \, (t+T+1)^{-1} \, dt $$
Recall $\Gamma (n+1) = n \Gamma (n) $. Therefore $\Gamma \big({2-\frac{2}{\alpha}}\big) = \big(1-\frac{2}{\alpha}\big) \Gamma \big({1-\frac{2}{\alpha}}\big)$
On further simplification and removing the exponents on LHS and RHS this reduces to
$$T\int_0^{\infty} t^{2} \, (t+1)^{{-1+\frac{2}{\alpha}}} \, (t+T+1)^{-1} \, dt = \Gamma\left ( \frac{2}{\alpha} \right )\Gamma\left ( 1-\frac{2}{\alpha } \right ) T^{\frac{2}{\alpha }} k $$
This is where I got stuck. For what value of $k$ do I proove that the above relations are equal