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After a fair bit of effort, I managed to prove that

$$\int_0^\infty t^\alpha \exp(-t) L_n^{\alpha+1}(t)\mathrm dt=\Gamma(\alpha+1)$$

where $L_n^\alpha (t)$ is a generalized Laguerre polynomial, with the usual restriction of $\Re \alpha > -1$. My proof was rather complicated (involving lots of integration by parts); I'm wondering if there may be a simpler way of proving this proposition.

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

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How about using the recurrence relation for generalized Laguerre polynomials?

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