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
$$\int_0^\infty dt\, t^\alpha e^{-t} L_n^{\alpha+1}(t) = \Gamma(\alpha+1) \qquad \forall n.$$