#### Answer

See proof below.

#### Work Step by Step

For a product, we have at least two factors (a product of one term makes no sense), so we prove that the formula is true for any $n\geq 2$
$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for $n=2$.
We are given by the problem text that
$f(x_{1}\cdot x_{2})=f(x_{1})+f(x_{2})$
$\qquad $
Yes, it is true when $n=2.$
$2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k$, THEN it also holds for the next integer, $n=k+1$.
Let us assume that for $n=k$,
$ f(x_{1}\cdot x_{2}\cdots\cdot x_{k})=f(x_{1})+f(x_{2})+\ldots+f(x_{k})$
(we call this the step hypothesis)
For $n=k+1$,
the LHS equals $f(x_{1}\cdot x_{2}\cdots\cdot x_{k}\cdot x_{k+1})$, which we rewrite as
$LHS= f[(x_{1}\cdot x_{2}\cdots\cdot x_{k})\cdot(x_{k+1})]$
$f $ of a product of TWO factors, for which we use the given formula
$=f(x_{1}\cdot x_{2}\cdots\cdot x_{k})+f(x_{k+1})$
the first term is, by the hypothesis, equal to $f(x_{1})+f(x_{2})+\ldots+f(x_{k})$
$=f(x_{1})+f(x_{2})+\ldots+f(x_{k})+f(x_{k+1})=RHS$ for $n=k+1$
which makes the formula true for $n=k+1.$
By mathematical induction, the formula is true for any integer $n\geq 2.$