Show that $E(\varepsilon)={\alpha}m$ when gamma distribution function is given as
$$f(\varepsilon)=\frac{1}{{\alpha^m\Gamma(m)}}\varepsilon^{m-1}\text{exp}[-\frac{\varepsilon}{\alpha}]$$
I am so stuck with calculations of this. I am using the formula of $\Gamma(\alpha+1)=\alpha\Gamma(\alpha)$ but I can't get the required result. I am stuck while counting the integral at the part where I would already plug in the limits. I get something like this before plugging in the limits: $\frac{m\Gamma(m)e^\frac{1}{\alpha}}{\alpha\Gamma(m)}$.