Here, it is useful to know the generating function ($|z|<1$) $ \frac{\exp\left(-\frac{xz}{1-z}\right)}{(1-z)^{\alpha+1}} = \sum_{n=0}^\infty L^\alpha_n (x) z^n .$ Thereby, we can evaluate the integral for all $n$ simultaneously $g=\sum_n z^n \int_0^\infty dt\, t^\alpha e^{-t} L_n^{\alpha+1}(t) = \int_0^\infty dt\, t^\alpha e^{-t} \frac{\exp\left(-\frac{tz}{1-z}\right)}{(1-z)^{\alpha+2}}. $ The integral can be brought onto the integral for the $\Gamma$ function by a change of variables $(1-z)^{\alpha+2} g= \int_0^\infty dt\, t^\alpha e^{ - \frac{t}{1-z} } = (1-z)^{\alpha+1} \int_0^\infty dt \, t^\alpha e^{ -t} =(1-z)^{\alpha+1} \Gamma(\alpha+1). $ Thus, we have $g = \Gamma(\alpha+1)/(1-z)$ and $\int_0^\infty dt\, t^\alpha e^{-t} L_n^{\alpha+1}(t) = \Gamma(\alpha+1) \qquad \forall n.$