Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 4 / 2C / Solution

The base case $n=1$ is the product rule. Suppose the Leibniz rule holds at order $n$. Taking one more derivative of each term and collecting the coefficient of $f^{(n+1-r)}{g}^{(r)}$ givesHere binomial coefficients outside $0\le r\le n$ are zero. For $1\le r\le n$, the factorial definition proves Pascal's identity directly:At $r=0,n+1$, the identity is $1+0=1$. Substitution gives the required order-$n+1$ formula, completing the induction.