Famous Theorems of Mathematics/e is transcendental

The mathematical constant is a transcendental number (or inalgebraic). In other words, it is not a root of any polynomial with rational coefficients. Proof Without loss of generality, it is sufficient to prove the claim for a polynomial with integer coefficients. Let us assume that is algebraic, so there exists a polynomial such that . Part 1 Let be a polynomial of degree . Let us define . Taking its derivative yields: Let us define . Taking its derivative yields: By the fundamental theorem of calculus, we get: Now let: Summing all the terms yields Part 2 Lemma: Let be a polynomial with a root of multiplicity . Then for all . Proof: By strong induction. Let us write , with a polynomial such that . For we get: Assume that for all the claim holds for all . We shall prove that for the claim holds for all : The blue blue part is of multiplicity , with a polynomial such that . Hence their product satisfies the induction hypothesis. Part 3 Let us now define a polynomial and is a prime number. We get: hence for all , the function is a polynomial with integer coefficients all divisible by . Now let us choose a prime . By part 2, for all we get: Therefore is also an integer divisible by . On the other hand, for we get: But , and are not divisible by . Therefore is not divisible by . Conclusion: is an integer not divisible by , and particularly . Hence . Part 4 By part 2, by the triangle inequality for integrals we get: By the triangle Inequality we get: But , hence for sufficiently large we get . A contradiction. Conclusion: is transcendental.