I'm following the book "Special functions of mathematical physics and chemistry" by Ian N. Sneddon where he derives Rodrigues formula for the Legendre polynomials
$$ P_n(\mu) = \frac{1}{2^n n!} \frac{d^n}{d\mu^n}(\mu^2 - 1)^n $$
from Murphy's formula
$$ P_n(\mu) = _2F _1 (-n,n+1;1;\frac{1-\mu}{2}) $$
Murphy's formula can be derived from the generating function of the Legendre polynomials (see Sneddon)
$$ (1-2\mu h + h^2)^{-\frac{1}{2}} = \sum_{n=0}^\infty P_n(\mu) h^n $$
My problem is that Sneddon derives the reader to the following exercise to complete the proof:
$$ _2 F _1 (-n, \alpha +n;\gamma;x) = \frac{x^{1-\gamma}(1-x)^{\gamma - \alpha}\Gamma(\gamma)}{\Gamma(\gamma+n)} \frac{d^n}{dx^n}(x^{\gamma +n-1}(1-x)^{\alpha - \gamma+n}) $$
And putting $\gamma \to \frac{\alpha}{2} + \frac{1}{2}$ and $ x \to \frac{1-\mu}{2}$ one can deduce
$$ _2 F _1 (-n, \alpha +n;\frac{\alpha + 1}{2};\frac{1-\mu}{2}) = \frac{(\mu^2 - 1)^{\frac{1 - \alpha}{2}}\Gamma(\frac{\alpha + 1}{2})}{2^n\Gamma(\frac{\alpha + 1}{2}+n)} \frac{d^n}{dx^n}((\mu^2 - 1)^{n + \frac{\alpha -1}{2}}) $$
Substituting here $\alpha = 1$ one derives Rodrigues formula because we already have proven Murphy's formula.
My question is how to prove the first of the latter two equations.
It follows from the general Leibniz rule and Pfaff's transformation.
To avoid confusion, I'm going to use $(a)_{n}$ for the falling factorial and $(a)^{(n)}$ for the rising factorial.
For $n \in \mathbb{N} $, the series representation for $_2 F_{1} (-n, \alpha+n;\gamma;x) $ is the finite series $$ \begin{align} _2 F_{1} (-n, \alpha+n;\gamma;x) &= \sum_{k=0}^{\infty} \frac{(-n)^{(k)} (\alpha+n)^{(k)}}{(\gamma)^{(k)}} \frac{x^{k}}{k!} \\ &=\sum_{k=0}^{n} (-1)^{k} \binom{n}{k}\frac{(\alpha+n)^{(k)}}{(\gamma)^{(k)}} x^{k}. \end{align} $$
Using the general Leibniz rule, we have
$$ \begin{align} &\frac{\partial^{n}}{\partial x^{n}} \left( x^{\gamma +n-1}(1-x)^{\alpha-\gamma+n} \right) \\ &= \sum_{k=0}^{n} \binom{n}{k} \frac{\partial^{k}}{\partial x^{k}} \left((1-x)^{\alpha-\gamma+n} \right) \frac{\partial^{n-k}}{\partial x^{n-k}} \left(x^{\gamma+n-1} \right) \\ &= \sum_{k=0}^{n} \binom{n}{k} (\alpha-\gamma+n)_{k} \, (1-x)^{\alpha-\gamma+n-k}(-1)^{k} \, (\gamma+n-1)_{n-k} \, x^{\gamma+n-1-(n+k)} \\ &\overset{(1)}{=} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k}(\gamma-\alpha-n)^{(k)}(1-x)^{\alpha-\gamma+n-k}(-1)^{k} \, (-1)^{n-k} (1-\gamma-n)^{(n-k)}x^{\gamma-1+k} \\ &= (-1)^{n} \sum_{k=0}^{n} (-1)^{k} \binom{n}{k}(\gamma-\alpha-n)^{(k)}(1-\gamma-n)^{(n-k)} x^{\gamma-1+k}(1-x)^{\alpha-\gamma+n-k} \\ &= (-1)^{n}x^{\gamma-1} (1-x)^{\alpha-\gamma+n} \sum_{k=0}^{n} (-1)^{k} \binom{n}{k}(\gamma-\alpha-n)^{(k)}(1-\gamma-n)^{(n-k)} \left(\frac{x}{1-x} \right)^{k} \\ &= (-1)^{n} x^{\gamma-1} (1-x)^{\alpha-\gamma+n} \sum_{k=0}^{n} \binom{n}{k}(\gamma-\alpha-n)^{(k)}(1-\gamma-n)^{(n-k)} \left(\frac{x}{x-1} \right)^{k} \\ &\overset{(2)}{=} (-1)^{n} x^{\gamma-1} (1-x)^{\alpha-\gamma+n} \sum_{k=0}^{n} \binom{n}{k}(\gamma-\alpha-n)^{(k)} \frac{(-1)^{n+k} (\gamma)^{(n)}}{(\gamma)^{(k)}} \left(\frac{x}{x-1} \right)^{k} \\ &= (\gamma)^{(n)}x^{\gamma-1} (1-x)^{\alpha-\gamma+n} \sum_{k=0}^{n} (-1)^{k} \binom{n}{k}(\gamma-\alpha-n)^{(k)} \frac{ 1}{(\gamma)^{(k)}} \left(\frac{x}{x-1} \right)^{k} \\ &= (\gamma)^{(n)} x^{\gamma-1} (1-x)^{\alpha-\gamma+n} \, _2F_{1} \left(-n, \gamma- \alpha -n; \gamma; \frac{x}{x-1} \right) \\ &\overset{(3)}{=} (\gamma)^{(n)} x^{\gamma-1} (1-x)^{\alpha-\gamma+n} (1-x)^{-n} \, _2F_{1} \left(-n, \alpha +n; \gamma; x \right) \\ &= \frac{\Gamma(\gamma+n)}{\Gamma(\gamma)}x^{\gamma-1} (1-x)^{\alpha-\gamma} \, _2F_{1} \left(-n, \alpha +n; \gamma; x \right). \end{align}$$
$(1)$ $(x)_{n}=(-1)^{n}(-x)^{(n)}$
$(2)$ $(1-\gamma-n)^{(n-k)} = \frac{\Gamma(1-\gamma-k)}{\Gamma(1-\gamma-n)} = \frac{\pi \csc \left(\pi(\gamma+k) \right)}{\Gamma(\gamma+k)} \frac{\Gamma(\gamma+n)}{\pi \csc \left(\pi(\gamma+n) \right)} \frac{\Gamma(\gamma)}{\Gamma(\gamma)} = (-1)^{n+k} \frac{(\gamma)^{(n)}}{(\gamma)^{(k)}}$
$(3)$ Pfaff's transformation in reverse