Polynomial conguence proof by GE Andrews

I shall appreciate views on the proof by GE Andrews in his "Number Theory" (Dover 1971) for Theorem 5-5: A polynomial with int coeff of degree n, whose leading coefficient is not divisible by the prime p, has at most n solutions modulo p.

First, I find Prof Andrew's book a fine introduction to the subject.

The proof is an indirect one and proceeds by mathematical induction on the degree n. After demonstration the cases for n=0,1, Andrews considers the polynomial f(x) of degree k with leading coefficiant, a, not divisible by p. The induction hypothesis is applied to polynomials of degree less than k. He constructs the polynomial g(x) from f(x) by subracting from f(x) the product, a * Pi(x - w). The w are solutions of f(x) = 0 (modp), of which there are posited k+1, contrary to the theorem - since f(x) is of deg k. The {w} of the product are any k of the k + 1 solutions modulo p. The difference between f(x) and the product, g(x), is of degree less than k. However g(x) has k solutions modulo p, as one may see! Thus, for g(x) the induction hypothesis has a false conclusion and, consequently, the antecedent of the hypothesis must be false as a matter of logic.

Herein lies my difficulty: the antecedent is a conjunction. Its denial can be had by the falsity of either conjunct. One conjunct is that its leading term is not divisible by p. That could be true or false. There is no specification of its divisibility. Only if it is not divisible by p are we forced to find that the other conjunct is false: that g(x)=0 (modp) for every x. Is the argument that since there is no specification as to the leading term of g(x) the conclusion must be that g(x) = 0 (modp) for every x ? That seems tenuous to me.

Also, I am not sure what the falsity of that conjunct really must mean. One could say that g(x) is identically 0, in which case g(x) = 0 (modp) is satisfied by every x.

Where am I missing the point, please ?