proposition: Lemma natf_neq0 n : (n%:R != 0 :> R) = (char R)^'.-nat n.
proof: Proof. have [-> | /prod_prime_decomp->] := posnP n; first by rewrite eqxx. rewrite !big_seq; elim/big_rec: _ => [|[p e] s /=]; first by rewrite oner_eq0. case/mem_prime_decomp=> p_pr _ _; rewrite pnatM pnatX eqn0Ngt orbC => <-. by rewrite natrM natrX mulf_eq0 expf_eq0 negb_or negb_and pnatE ?inE p_pr. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop prime binomial)