I am studying parametric statistical inference. One of the self study I have to find a sufficient, minimal and complete statistic for the $\mu$ parameter of the following p.d.f. $$ f_X(x \mid \mu) = e^{-(x - \mu)} I_{(\mu, \infty)}(x) $$ which is a exponential distribution with location parameter $\mu$.
We can write the p.d.f. of a random sample $\mathbf{x} = (x_1, \ldots, x_n)$ of $X$ as $$ f_{\mathbf{X}}(\mathbf{x} \mid \mu) = e^{-n(\bar{x} - \mu)}\,I_{(\mu, \infty)}(x_{(1)}) $$ where $x_{(1)} = \min(\mathbf{x})$.
By the Factorization Theorem I concluded that $T = X_{(1)}$ is a sufficient statistic for $\mu$. To prove that $X_{(1)}$ is also minimal I showed that the ratio $f_{\mathbf{X}}(\mathbf{x} \mid \mu)/f_{\mathbf{X}}(\mathbf{y} \mid \mu)$ does not depend on $\mu$ iff $x_{(1)} = y_{(1)}$
In regard the completeness of $X_{(1)}$ I have to prove that $E(g(T)) = 0$ for all $\mu$, i.e., there is no function of $T = X_{(1)}$ unless the $g(T) = 0$ zero function.
I have found the distribution of $T$, which is given by $$ f_T(t) = n\,e^{-n(x-\mu)} I_{(\mu, \infty)}(t). $$$$ f_T(t) = n\,e^{-n(t-\mu)} I_{(\mu, \infty)}(t). $$
However, I couldn't show that $E(g(T)) = 0$.
Is there another way to prove that $T$ is or is not a complete statistic for $\mu$?