How to use induction to prove the product rule for higher derivatives? How do I show through mathematical induction that 
\begin{equation}
\frac{d^n}{dx^n}[f(x)\cdot g(x)] = \sum_{k = 0}^{n} \binom{n}{k} \frac{d^k}{dx^k} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)]
\end{equation}
which is the product rule for higher derivatives?
 A: I assume you already know the base case, which is the standard product rule:

\begin{equation}
\frac{d}{dx}[f(x)\cdot g(x)] =  \frac{d}{dx} [f(x)] \cdot  g(x) + f(x) \cdot \frac{d}{dx}[g(x)]
\end{equation}

If you don't, here are two links for the necessary prerequisites:
Product Rule: https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/productruledirectory/ProductRule.html
Proof by Induction: http://www.purplemath.com/modules/inductn.htm
So now we need to state the induction hypothesis:

\begin{equation}
\frac{d^{n-1}}{dx^{n-1}}[f(x)\cdot g(x)] = \sum_{k = 0}^{n-1} \binom{n-1}{k} \frac{d^k}{dx^k} [f(x)] \cdot \frac{d^{n-1-k}}{dx^{n-1-k}} [g(x)]
\end{equation}

and then prove that the induction step

\begin{equation}
\frac{d^n}{dx^n}[f(x)\cdot g(x)] = \sum_{k = 0}^{n} \binom{n}{k} \frac{d^k}{dx^k} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)]
\end{equation}

holds, i.e. is true.
Allow us to apply the standard product rule onto the induction hypothesis and see what we find:

\begin{equation}
\frac{d}{dx} \left[\frac{d^{n-1}}{dx^{n-1}}[f(x)\cdot g(x)]\right] = \frac{d}{dx} \left[ \sum_{k = 0}^{n-1} \binom{n-1}{k} \frac{d^k}{dx^k} [f(x)] \cdot \frac{d^{n-1-k}}{dx^{n-1-k}} [g(x)] \right] \\ = \sum_{k = 0}^{n-1} \binom{n-1}{k} \frac{d}{dx} \left[ \frac{d^k}{dx^k} [f(x)] \cdot \frac{d^{n-1-k}}{dx^{n-1-k}} [g(x)] \right] \quad \text{by linearity of the derivative} \\ = \sum_{k = 0}^{n-1} \left[ \binom{n-1}{k}  \left[ \frac{d^{k+1}}{dx^{k+1}} [f(x)] \cdot \frac{d^{n-(k+1)}}{dx^{n-(k+1)}} [g(x)] \right] + \binom{n-1}{k} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right] \right]
\end{equation}

the last equality following by the standard product rule.
We now split the last sum into two sums

\begin{equation}
\sum_{k=0}^{n-1}  \binom{n-1}{k}  \left[ \frac{d^{k+1}}{dx^{k+1}} [f(x)] \cdot \frac{d^{n-(k+1)}}{dx^{n-(k+1)}} [g(x)] \right] + \sum_{k=0}^{n-1} \binom{n-1}{k} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right]
\end{equation}

Reindexing the left sum we get

\begin{equation}
\sum_{k=1}^{n-1}  \binom{n-1}{k-1}  \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right]+ \frac{d^n}{dx^n}f(x) \cdot g(x) + \sum_{k=1}^{n-1} \binom{n-1}{k} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right] + f(x) \cdot \frac{d^n}{dx^n} g(x)
\end{equation}

We now use the Pascal rule for binomial coefficients

$$ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$

and combine the sums again for $k=1,\dots,n-1$, the values of $k$ which the two have in common:

\begin{equation}
\sum_{k=1}^{n-1} \binom{n}{k} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right] + \frac{d^n}{dx^n}f(x) \cdot g(x) + f(x) \cdot \frac{d^n}{dx^n} g(x) 
\end{equation}

This is equal to 

\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} \left[ \frac{d^{k}}{dx^{k}} [f(x)] \cdot \frac{d^{n-k}}{dx^{n-k}} [g(x)] \right]
\end{equation}

Which is what we wanted to show.
