Proof by induction: === Basis for the Induction === Let $n = 1$. From Product Rule for Derivatives: :$\paren {\map f x \, \map g x}' = \paren {\map f x \, \map {g'} x} + \paren {\map {f'} x \, \map g x}$ Likewise: :$\displaystyle \sum_{k \mathop = 0}^1 \binom 1 k \map {f^{\paren k} } x \, \map {g^{\paren {1 - k} } } x = \binom 1 0 \map f x \, \map {g^{\paren {1 - 0} } } x + \binom 1 1 \map {f'} x \, \map {g^{\paren {1 - 1} } } x = \map f x \, \map {g'} x + \map {f'} x \, \map g x$ This is our basis for the induction. === Induction Hypothesis === Let $n \in \N$ be fixed. We assume the inductive hypothesis: :$\displaystyle \paren {\map f x \, \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x$ We need to show that: :$\displaystyle \paren {\map f x \, \map g x}^{\paren {n + 1} } = \sum_{k \mathop = 0}^{n + 1} \binom {n + 1} k \map {f^{\paren k} } x \, \map {g^{\paren {n + 1 - k} } } x$ === Induction Step === By our inductive hypothesis: {{begin-eqn}} {{eqn | l = \paren {\map f x \, \map g x}^{\paren {n + 1} }       | r = \paren {\sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x}' }} {{eqn | r = \sum_{k \mathop = 0}^n \binom n k \paren {\paren {\map {f^{\paren k} } x \map {g^{\paren {n - k} } } x}'}       | c = Applications of Linear Combination of Derivatives }} {{eqn | r = \sum_{k \mathop = 0}^n \binom n k \paren {\map {f^{\paren {k + 1} } } x \map {g^{\paren {n - k} } } x            + \map {f^{\paren k} } x \, \map {g^{\paren {n + 1 - k} } } x}       | c = from the base case }} {{eqn | r = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren {k + 1} } } x \, \map {g^{\paren {n - k} } } x             + \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n + 1 - k} } } x       | c = splitting the summation }} {{end-eqn}} Subsequently, we separate the $k = 0$ case from the second summation.  For the first summation, we separate the case $k = n$ and then shift the indices up by $1$. These manipulations give us the following: {{begin-eqn}} {{eqn | l = \paren {\map f x \, \map g x}^{\paren {n + 1} }       | r = \sum_{k \mathop = 1}^n \binom n k \map {f^{\paren k} } x \map {g^{\paren {n + 1 - k} } } x           + \sum_{k \mathop = 1}^n \binom n {k - 1} \map {f^{\paren k} } x \, \map {g^{\paren {n + 1 - k} } } x           + \binom n 0 \map f x \map {g^{\paren {n + 1} } } x           + \binom n n \map {f^{\paren {n + 1} } } x \, \map g x }} {{end-eqn}} By Pascal's Rule, we finally obtain: {{begin-eqn}} {{eqn | l = \paren {\map f x \, \map g x}^{\paren {n + 1} }       | r = \sum_{k \mathop = 1}^n \binom {n + 1} k \map {f^{\paren k} } x \, \map {g^{\paren {n + 1 - k} } } x           + \binom n 0 \map f x \, \map {g^{\paren {n + 1} } } x           + \binom n n \map {f^{\paren {n + 1} } } x \map g x  }} {{eqn | r = \sum_{k \mathop = 1}^n \binom {n + 1} k \map {f^{\paren k} } x \, \map {g^{\paren {n + 1 - k} } } x           + \binom {n + 1} 0 \map f x \map {g^{\paren {n + 1} } } x           + \binom {n + 1} {n + 1} \map {f^{\paren {n + 1} } } x \, \map g x  }} {{eqn | r = \sum_{k \mathop = 0}^{n + 1} \binom {n + 1} k \map {f^{\paren k} } x \, \map {g^{\paren {n + 1 - k} } } x  }} {{end-eqn}} The result follows by the Principle of Mathematical Induction. {{qed}}