Theorem: Gauss's Hypergeometric Theorem

:$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$

Proof:
Let $x, y, n \in \C$ be complex numbers such that $\map \Re {x + y + n + 1} > 0$.
Let $u \in \C$ be a complex number such that $\cmod u < 1$.
Expanding the product of $\paren {1 + u}^{y + n}$ and $\paren {\dfrac {1 + u} u}^x$:
{{begin-eqn}}
{{eqn | l = \paren {1 + u}^{y + n}
      | r = \sum_{k \mathop = 0}^\infty \binom {y + n} k u^k
      | c = Binomial Theorem - Complex Numbers
}}
{{eqn | l = \paren {1 + \dfrac 1 u}^x
      | r = \sum_{k \mathop = 0}^\infty \binom x k u^{-k}
      | c = Binomial Theorem - Complex Numbers, Power of Product
}}
{{eqn | ll= \leadsto
      | l = \paren {1 + u}^{y + n} \paren {\dfrac {1 + u} u}^x
      | r = \sum_{k \mathop = 0}^\infty \binom {y + n} k u^k \sum_{k \mathop = 0}^\infty \binom x k u^{-k}
      | c = multiplying
}}
{{end-eqn}}
The coefficient $a_n$ of $u^n$ of the product $\dfrac {\paren {1 + u}^{x + y + n} } {u^x}$ above can be determined by setting $k = k + n$ in the series with $\dbinom {y + n} k$:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \sum_{k \mathop = 0}^\infty \binom {y + n} {k + n} \binom x k
      | c = substituting $k = k + n$ in first series: $u^n = u^{k+n} u^{-k}$
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {y + n}!} {\paren {k + n}! \paren {y - k}!} \dfrac {x!} {k! \paren {x - k}!}
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {y + n}!} {\paren {k + n}! \paren {y - k}!} \dfrac {x!} {k! \paren {x - k}!} \paren {\dfrac {n! y!} {n! y!} }
      | c = multiplying by $1$
}}
{{eqn | r = \dfrac {\paren {y + n}! } {n! y!}  \sum_{k \mathop = 0}^\infty \dfrac {n! } {k! \paren {k + n}! } \dfrac {y!} {\paren {y - k}!}  \dfrac {x!} {\paren {x - k}!} 
      | c = moving $\dfrac {\paren {y + n}! } {n! y!}$ outside the sum and rearranging
}}
{{eqn | r = \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} }  \sum_{k \mathop = 0}^\infty \dfrac {n! } {k! \paren {k + n}! } \dfrac {y!} {\paren {y - k}!} \dfrac {k!} {k!}  \dfrac {x!} {\paren {x - k}!} \dfrac {k!} {k!}
      | c = multiplying by $1$
}}
{{eqn | r = \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} }  \sum_{k \mathop = 0}^\infty \dfrac {n! } {k! \paren {k + n}! } \dbinom y k \dbinom x k \paren {k!}^2
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} }  \sum_{k \mathop = 0}^\infty \dfrac {n! } {k! \paren {k + n}! } \paren {-1}^k \dbinom {-y + k - 1} k \paren {-1}^k \dbinom {-x + k - 1} k \paren {k!}^2
      | c = Negated Upper Index of Binomial Coefficient
}}
{{eqn | r = \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} }  \sum_{k \mathop = 0}^\infty \dfrac {n! } {k! \paren {k + n}! } \dfrac {\paren {-y + k - 1}!} {k! \paren {-y - 1}!} \dfrac {\paren {-x + k - 1}!} {k! \paren {-x - 1}!} \paren {k!}^2
      | c = {{Defof|Binomial Coefficient}}, $\paren {-1}^{2 k} = 1$
}}
{{eqn | r = \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} } \sum_{k \mathop = 0}^\infty \dfrac { \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { k! \paren {n + 1}^{\overline k} }
      | c = Rising Factorial as Quotient of Factorials, and $k!$ cancels
}}
{{end-eqn}}
Now expand $\paren {1 + u}^{x + y + n}$ and divide by $u^x$:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {1 + u}^{x + y + n} } {u^x}
      | r = \sum_{k \mathop = 0}^\infty \binom {x + y + n} k u^{k - x}
      | c = Binomial Theorem - Complex Numbers
}}
{{end-eqn}}
The coefficient $a_n$ of $u^n$ of the product $\dfrac {\paren {1 + u}^{x + y + n} } {u^x}$ above is:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \binom {x + y + n} {x + n}
      | c = $n = k - x$, so $k = x + n$
}}
{{eqn | r = \dfrac {\map \Gamma {x + y + n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + 1} }
      | c = {{Defof|Binomial Coefficient}}
}}
{{end-eqn}}
Equating coefficients gives us:
{{begin-eqn}}
{{eqn | l = \dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} } \sum_{k \mathop = 0}^\infty \dfrac { \paren {-x}^{\overline k} \paren {-y}^{\overline k} } {\paren {n + 1}^{\overline k} } \dfrac 1 {k!}
      | r = \dfrac {\map \Gamma {x + y + n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 0}^\infty \dfrac { \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { \paren {n + 1}^{\overline k} } \dfrac 1 {k!}
      | r = \dfrac {\map \Gamma {x + y + n + 1} \map \Gamma {n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} }
      | c = dividing both sides by $\dfrac {\map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {y + 1} }$
}}
{{eqn  | ll= \leadsto
      | l = \map F {-x, -y; n + 1; 1}
      | r = \dfrac {\map \Gamma {x + y + n + 1} \map \Gamma {n + 1} } {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} }
      | c = {{Defof|Hypergeometric Function}}
}}
{{end-eqn}}
Letting $a = -x$, $b = -y$ and $c = n + 1$, we obtain:
{{begin-eqn}}
{{eqn | l = \map F {a, b; c; 1}
      | r = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }
      | c = 
}}
{{end-eqn}}
{{qed}}