There is a theorem that says that the convergent exponent of the zeros of a entire function is never greater than its order, so i'm traying to use that.
The zeros of the reciprocal gamma function are the poles of the $\Gamma$, thatss form se sequence $\{a_{n}\}$ $\{0,-1,-2,-3,...\}$ and the exponent $\lambda$ is the greatest lower bound such that
$$\sum_{n=1}^{\infty}\frac{1}{|a_{n}|^{\lambda}}$$
converges, well we all know thats form the $\lambda$-harmonic series that converges for $\lambda >1$, so all the exponents $p=1+\epsilon, \forall \epsilon >0$ is a lower bound for the order $\rho$ of $f$, but if $p=1$ the serie diverges so how can i conclude that $\rho =1$?