that all but at most one of them are transcendental. In 2012 it was shown that at least one of and the Euler–Gompertz constant is transcendental. Apéry's constant (which Apéry proved is irrational). Catalan's constant, not even proven to be irrational. Khinchin's constant, also not proven to be irrational. The Riemann zeta function at other odd integers, , , ... (not proven to be irrational). The Feigenbaum constants and , also not proven to be irrational. Mills' constant, also not proven to be irrational. The Copeland–Erdős constant, formed by concatenating the decimal representations of the prime numbers. has not been proven to be irrational. Conjectures: Schanuel's conjecture, Four exponentials conjecture. Sketch of a proof that is transcendental The first proof that the base of the natural logarithms, , is transcendental dates from 1873. We will now follow the strategy of David Hilbert (1862–1943) who gave a simplification of the original proof of Charles Hermite. The idea is the following: Assume, for purpose of finding a contradiction, that is algebraic. Then there exists a finite set of integer coefficients c0, c1, ..., cn satisfying the equation: Now for a positive integer k, we define the following polynomial: and multiply both sides of the above equation by to arrive at the equation: By splitting respective domains of integration, this equation can be written in the form whereLemma 1. For an appropriate choice of k, is a non-zero integer. Proof. Each term in P is an integer times a sum of factorials, which results from the relation which is valid for any positive integer j (consider the Gamma function). It is non-zero because for every a satisfying 0< a ≤ n, the integrand in is e−x times a sum of terms whose lowest power of x is k+1 after substituting x for x+a in the integral. Then this becomes a sum of integrals of the form Where Aj-k is integer. with k+1 ≤ j, and it is therefore an integer divisible by (k+1)!. After dividing by k!, we get zero modulo (k+1). However, we can write: and thus So when dividing each integral in P by k!, the initial one is not divisible by k+1, but all the others are, as long as k+1 is prime and larger than n and |c0|. It follows that itself is not divisible by the prime k+1 and therefore cannot be zero.Lemma 2.''' for sufficiently large .