When writing the relation for the Legendre polynomials $$ P^{-\mu}_{\nu}(z) = (-1)^{\mu} {\Gamma(\nu-\mu+1) \over \Gamma(\nu+\mu+1)} P^{\mu}_{\nu}(z) $$ in the Gauss hypergeometric representation, we obtain $$ F\left( -\nu, \nu+1,1+\mu,{1-z \over 2} \right) = (-1)^{\mu} {\Gamma(\nu-\mu+1) \over \Gamma(\nu+\mu+1)} {\Gamma(1+\mu) \over \Gamma(1-\mu)} \left({1+z \over 1-z}\right)^{\mu} F\left( -\nu, \nu+1,1-\mu,{1-z \over 2} \right)\,. $$ My question is, how could one find this relation using only the properties of hypergeometric functions? Every manipulation I try using transformations between hypergeometric functions always result in 2 terms.
EDIT: Corrected the last equation.
EDIT2: I have managed to get the following steps
\begin{align} & \left({1+z \over 2} \right)^{-\mu} F\left[ -\nu,\nu+1;1+\mu; {1-z \over 2} \right] = \\ & = F\left[ 1+\mu+\nu,\mu-\nu;1+\mu; {1-z \over 2} \right] \\ & = \sum^{\infty}_{k=0} {\Gamma\left(1+\mu+\nu + k\right) \over \Gamma\left(1+\mu+\nu\right)}{\Gamma\left(\mu-\nu + k\right) \over \Gamma\left(\mu-\nu\right)}{\Gamma\left(1+\mu\right) \over \Gamma\left(1+\mu+k\right)} {1\over k!} \left({1-z \over 2}\right)^k \\ & = {\Gamma\left(1+\mu\right) \over \Gamma\left(1-\mu\right)} {\Gamma\left(1+\nu\right)\Gamma\left(-\nu\right) \over \Gamma\left(1+\mu+\nu\right)\Gamma\left(\mu-\nu\right)}\left({1-z \over 2}\right)^{-\mu} \times \sum^{\infty}_{k'=\mu} {\Gamma\left(1+\nu + k'\right) \over \Gamma\left(1+\nu\right)}{\Gamma\left(-\nu + k'\right) \over \Gamma\left(-\nu\right)}{\Gamma\left(1-\mu\right) \over \Gamma\left(1-\mu+k'\right)} {1\over k'!} \left({1-z \over 2}\right)^{k'} \\ & = {\Gamma\left(1+\mu\right) \over \Gamma\left(1-\mu\right)} (-1)^{\mu}{\Gamma\left(1-\mu+\nu\right) \over \Gamma\left(1+\mu+\nu\right)}\left({1-z \over 2}\right)^{-\mu} \sum^{\infty}_{k'=\mu} { (-\nu)_{k'} (\nu+1)_{k'} \over (1-\mu)_{k'} } {1\over k'!} \left({1-z \over 2}\right)^{k'} \end{align}
where I defined $k' = k+\mu$ and used the reflection formula to simplify the $\Gamma$ functions. The very last step would be to somehow turn the summation from 0 to $\infty$ rather than $\mu$ to $\infty$, but I cannot see how this could be done.