The derivative of a product of more than two functions

Simplest way to establish this is by induction on $n$.

The case $n=1$ is immediate; the case $n=2$ is the usual product rule. Assuming you have established the desired formula $$\left(\prod_{i=1}^n f_i(x)\right)' = \sum_{i=1}^n \left(f_i'(x)\prod_{\stackrel{1\leq j\leq n}{i\neq j}}f_j(x)\right)$$ for $n$, then to get the $n+1$ case we have: $$\begin{align*} \left(\prod_{i=1}^{n+1}f_i(x)\right)' &= \left(\left(\prod_{i=1}^n f_i(x)\right)f_{n+1}(x)\right)'\\ &= \left(\prod_{i=1}^nf_i(x)\right)'f_{n+1}(x) + \left(\prod_{i=1}^nf_i(x)\right)f_{n+1}^{'}(x)\\ &= \left(\sum_{i=1}^nf_i'(x)\prod_{\stackrel{1\leq j\leq n}{i\neq j}}f_j(x)\right)f_{n+1}(x) + \left(\prod_{i=1}^nf_i(x)\right)f_{n+1}'(x)\\ &= \sum_{i=1}^nf_i'(x)\prod_{\stackrel{1\leq j\leq n+1}{i\neq j}}f_j(x) + \left(\prod_{i=1}^nf_i(x)\right)f_{n+1}'(x)\\ &=\sum_{i=1}^{n+1} f_i'(x)\prod_{\stackrel{1\leq j\leq n+1}{i\neq j}}f_j(x), \end{align*}$$ as desired.

You can use induction on $n$, the number of functions. if $n = 1$, there is nothing to prove. if $n = 2$, then you just get the product rule. Assume the claim is true for $n$ functions, and prove it for $n+1$. Write $f_1f_2...f_{n+1}$ = $f_1g$ where $g = f_2..f_{n+1}$. Now differentiate $f_1g$ using the product rule and apply the induction hypothesis to $g'$. Note that $g$ is a product of $n$ functions, so the induction hypothesis tells you what $g'$ is.