We present a constructive proof that P ̸= NP. The core of the argument is a factorial growthrecurrence that models the minimal number of logical distinctions any exact deterministic solver for SAT must perform. We show that the unique solution of this recurrence is f (n!) = n!, forcing factorial-time complexity. Crucially, we prove that this recurrence is universal: any deterministic Turing machine deciding SAT exactly must implicitly realize this factorial explosion, regardless of its algorithmic design. We reinforce this conclusion with an algebraic irreducibility argument: 3SAT clauses, represented by cubic Boolean polynomials, cannot be reduced to quadratic (2SAT) constraints without exponential overhead. This demonstrates that no encoding or algebraic compression can avoid the factorial structure.Finally, we discuss why this approach circumvents classical barriers (relativization, natural proofs, algebrization). Together, these results yield a universal contradiction with the P = NP hypothesis. We conclude that P ̸= NP.