=$\displaystyle \int_{0}^{\infty} e^\frac{y(t\theta-1)}{\theta} \frac{y^{n-1}}{\theta^{n-1}}\frac{1}{\theta(n-1)!}dy$

let u=$\displaystyle {\frac{y(t\theta-1)}{\theta}} $

Then du=$\displaystyle \frac{(t\theta-1)}{\theta}$dy

=$\displaystyle \int_{0}^{\infty} \frac{1}{(t\theta-1)(n-1)!} \frac{u^{n-1}}{(t\theta-1)^{n-1}}e^u du$

= $\displaystyle \frac{1}{(t\theta-1)^n}$ $\displaystyle \int_{0}^{\infty} e^u \frac{u^{n-1}}{(n-1)!} du$

Is this right so far? Is there some reason why $\displaystyle \int_{0}^{\infty} e^u \frac{u^{n-1}}{(n-1)!} du$ should be one? Could somebody show me how to integrate this by parts? (I know I should be able to do it but I am really struggling).