I think you can resolve this problem by making the construction a bit more explicit. Suppose that (terms of lower degree). Suppose also that the equation f(x) = 0 (mod p) has solutions (mod p). Then we form the polynomial , which has leading term .

As I understand it, your difficulty with George Andrews's proof stems from the fact that the coefficient of the leading term in the polynomial g(x) might be divisible by p, in which case the inductive proof fails to achieve its aim. That looks like a very valid objection to me. But one can easily get round it by exchanging one of the s with so as to change the value of the product in that leading coefficient, which will then no longer be a multiple of p. That will remove the undesired element of the conjunction from the inductive hypothesis, so that the proof can be successfully completed.