Proving Saalschutz Theorem I saw this in a pdf, and I'm wondering

Questions:

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*How do you prove Saalschutz Theorem:

$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\dfrac {\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+x+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}\tag{1}$$

I'm somewhat relatively new to Hypergeometrical Series. I understand that the general Hypergeometrical series takes the form$$_pF_q\left[\begin{array}{c,c}\alpha_1,\alpha_2,\ldots,\alpha_p\\\beta_1,\beta_2,\ldots,\beta_q\end{array};x\right]=\sum\limits_{k=0}^{\infty}\dfrac {(\alpha_1)_k(\alpha_2)_k\ldots(\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\ldots(\beta_q)_k}\dfrac {x^k}{k!}\tag{2}$$
So therefore, by $(2)$, we should have$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\sum\limits_{k=0}^{\infty}\dfrac {(-x)_k(-y)_k(-z)_k}{(n+1)_k(-x-y-z-n)_k}\tag{3}$$
However, I'm not sure how to manipulate the RHS of $(3)$ to get the RHS of $(1)$.
EDIT: Since $(a)_k=\Gamma(a+k)/\Gamma(a)$, the RHS of $(3)$ becomes$$\dfrac {(-x)_k(-y)_k(-z)_k}{(n+1)_k(-x-y-z-n)_k}=\dfrac {\Gamma(k-y)\Gamma(n+1)\Gamma(k-x)\Gamma(-x-y-z-n)\Gamma)k-z)}{\Gamma(n+k+1)\Gamma(-x)\Gamma(-y)\Gamma(-z)\Gamma(-x-y-z-n+k)}$$Now, I need to figure out how$$\Gamma(k-y)\Gamma(k-x)\Gamma(k-z)\Gamma(-x-y-z-n)=\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+z+n+1)$$$$\Gamma(n+k+1)\Gamma(-x)\Gamma(-y)\Gamma(-z)\Gamma(-x-y-z-n+k)=\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)$$


Extra: I also believe that using the same general approach, we can prove$$\begin{align*} & _7F_6\left[\begin{array}{c,c}n,\frac 12n+1,-x,-y,-z,-u,x+y+z+u+2n+1\\\frac 12n,x+n+1,y+n+1,z+n+1,u+n+1,-x-y-z-u-n\end{array}\right]\\ & =\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(u+n+1)\Gamma(x+y+z+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+u+n+1)\Gamma(z+u+n+1)}\\ & \times\dfrac {\Gamma(y+z+u+n+1)\Gamma(x+u+z+n+1)\Gamma(x+y+u+n+1)}{\Gamma(x+z+n+1)\Gamma(y+u+n+1)\Gamma(x+y+z+u+n+1)}\end{align*}\tag{4}$$



Prove$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\dfrac {\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+x+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}$$


Proof: Begin with the identity$$(1-z)^{a+b-c}\space_2F_1(a,b;c;z)=_2F_1(c-a,b-a;c;z)\tag1$$This can be easily proven by setting both solutions of the second order linear differential equation$$(z-z^2)\frac {d^2y}{dz^2}+\bigr\{c-(a+b+1)z\bigr\}\frac {dy}{dx}-aby=0$$equal to each other, and changing the dependent variables. Starting with $(1)$, rewrite it as a summation, and then find the coefficient of $z^n$.$$\sum\limits_{k=0}^\infty\frac {(a+b-c)_k}{k!}(-z)^k\sum\limits_{r=0}^{\infty}\frac {(a)_r(b)_r}{(c)_r}\frac {z^r}{r!}=\sum\limits_{l=0}^{\infty}\frac {(c-a)_l(b-a)_l}{(c)_l}\frac {z^l}{l!}\tag2$$The coefficient of $z^n$ of $(2)$ is therefore$$\sum\limits_{r=0}^{\infty}\frac {(a)_r(b)_r(c-a-b)_{n-r}}{(n-r)!(c)_rr!}=\frac {(c-a)_n(b-a)_n}{(c)_nn!}\tag3$$And from $(3)$, it follows that the left-hand side is equal to$$\sum\limits_{r=0}^{\infty}\frac {(a)_r(b)_r}{(c)_r}\frac {(c-a-b)_n(-n)_r}{(1+a+b-c-n)_rn!}=_3F_2\left[\begin{array}{c c}a,b,-n\\c,1+a+b-c-n\end{array}\right]\frac 1{n!(c-a-b)_n}$$Equating to the right-hand side of $(3)$, and simplifying, we get the identity$$_3F_2\left[\begin{array}{c c}a,b,-n\\c,1+a+b-c-n\end{array}\right]=\frac {\Gamma(c)\Gamma(c-a-b)\Gamma(n-a+b)\Gamma(n-a+c)}{\Gamma(b-a)\Gamma(c-a)\Gamma(c+n)\Gamma(n+c-a-b)}\tag4$$Replacing $a=-x,b=-y,n=z,$ and $c$ with $n+1$, we deduce Saalschutz's theorem.
