A congruence modulo p^2 for the p-th coefficient of e(1-z)^((1-z)/z)

Define E_n by sum_{n>=0} E_n z^n = exp(sum_{k>=1} z^k/(k(k+1))) = e(1-z)^((1-z)/z). Equivalently, e(1+1/x)^(-x) = sum_{n>=0} E_n (x+1)^(-n), the reciprocal of the expansion of e^(-1)(1+1/x)^x used by Yang in refinements of Carleman's inequality. We prove that for every odd prime p, E_p = -p/2 (mod p^2); in particular v_p(E_p) = 1. The proof is elementary. With Phi(u) = e(1+u)^(-1/u), it combines the two expressions of E_p through the coefficients of Phi that come from the identities e(1-z)^((1-z)/z) = Phi(-z)/(1-z) and e(1-z)^((1-z)/z) = Phi(z/(1-z)), and it reduces the congruence to the differential recurrence of the E_n read modulo p. A script checks the statement and the intermediate congruences for all odd primes below 200 and a few larger ones.