A careful proof will be a maze of notation, and probably not helpful. Perhaps a sketch will help.
f is the sum of \Delta(f) terms, each of the form a x^{r(p-1)}.
g is the sum of \Delta(g) terms, each of the form b x^s, with s<p-1.
When calculating fg, we multiply each of the \Delta(f) terms by each of the \Delta(g) terms, and add the products together.
How many terms are there in fg? Certainly no more than \Delta(f)\Delta(g). But could there be less? No, because each of the products is non-zero, and all the powers of the resulting x^{r(p-1)+s} are different.
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How can a coeffiecient be 0?
For convinience lets say q=p-1.
See f is a polynomial with terms of multiple of power q,g is a polynomial with degree<q;
Lets say i is the power of a term in result,so i will be (j*q)+k where j denotes x^(jq) power if present in f,k denotes x^k power if present in g;
So as u can see ,i will always be distinct,so for each mul we get a distinct term,so for the coefficient to be 0,(a*b)modp should be 0(a is coefficient in f,b in g),which is not possible if a!=0,b!=0,p->prime
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oh sorry!! i get it now. This holds true because p is prime so, a*b mod p is never equal to 0. Thanks a lot!
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