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The generalized Leibniz rule follows by mathematical induction. The case $n=0$ is the identity $uv=uv$. Suppose the formula holds for $n$ and differentiate every term using the product rule:For an interior index $r$, the coefficient of $u^{(r)}{v}^{(n+1-r)}$ becomes $\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)+\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$ by Pascal's identity. The two endpoint coefficients are one. This provesOnly finitely many derivatives are involved; it suffices that both functions are $n$ times differentiable.

For $u=x(x+1)$ and $v=e^{2x}$, the derivatives of $u$ beyond order two vanish. The three surviving terms of the generalized Leibniz rule giveThe simplified expression works for every integer $n\ge 0$, including the two cases where one or more of the three terms are absent.

For $u=x^2$ and $v=\sin x$, use $d^m(\sin x)/d{x}^m=\sin (x+m\pi /2)$. Again only three terms survive. With $T=x+n\pi /2$, they areSince $\sin (T-\pi /2)=-\cos T$ and $\sin (T-\pi )=-\sin T$, this becomesThe factors $n$ and $n(n-1)$ remove the nonexistent terms when $n=0$ or $1$.

For functions with the required derivatives, the Leibniz rule isAt $n=0$ it is immediate. If it holds at $n$, differentiating each summand gives two sums. The coefficient of $f^{(k)}{g}^{(n+1-k)}$ after regrouping is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)+\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)$ by Pascal's identity, including the two endpoint terms. This proves the formula by mathematical induction.

Apply it to $(fg)h$ and then expand the derivatives of $fg$. For $p+q+r=n$, the resulting coefficient is $\left(\genfrac{}{}{0ex}{}{n}{p+q}\right)\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)=n!/(p!q!r!)$. Consequentlywhere $p,q,r$ range over nonnegative integers. This is the three-factor generalized Leibniz rule.