Just take
>S := x1 x2 x3 ... xk
Note that under your assumption (f holds for k),
>f(S) = f(x1) + f(x2) ... + f(xk)

Consider the series
>f(x1) + f(x2) ... + f(xk) + f(x[k+1])
We see that this is equal to
>f(S) + f(x[k+1])
Which by the definition of f is
>f(S x[k+1])
Substitute in the definition of S to get
>f(x1 x2 x3 ... xk x[k+1])
So
>f(x1 x2 x3 ... xk x[k+1]) =f(x1) + f(x2) ... + f(xk) + f(x[k+1])

Thus, by induction (omitting the proof that f is defined for n=1)
f(x1 x2 x3 ... xn) = f(x1) + f(x2) + ... + f(xn) for n>0 (zero), n is an integer.