TITLE: Deriving Laplace Transform of Laguerre polynomial
QUESTION [7 upvotes]: I'm given this definition for the Laguerre polynomials:
$$L_n(t)=\frac{e^t}{n!}\frac{d^n}{dt^n}\left[t^ne^{-t}\right],~\text{for }n=0,1,2...$$
and I have to show that the Laplace transform is
$$\frac{1}{s}\left(\frac{s-1}{s}\right)^n$$
My first attempt was to find a formula for the $n$th derivative of $t^ne^{-t}$, which I found to be
$$\frac{d^n}{dt^n}\left[t^ne^{-t}\right]=e^{-t}\sum_{k=0}^n\frac{n!}{(n-k)!}\binom nk (-t)^{n-k}$$
So from here I tried taking the transform:
$$\begin{align*}\mathcal{L}_s\left\{\frac{e^t}{n!}\frac{d^n}{dt^n}\left[t^ne^{-t}\right]\right\}&=\frac{1}{n!}\mathcal{L}_{s-1}\left\{e^{-t}\sum_{k=0}^n\frac{n!}{(n-k)!}\binom nk (-t)^{n-k}\right\}\end{align*}$$
where $\mathcal{L}_s\{f(t)\}=F(s)$ and $\mathcal{L}_s\{e^tf(t)\}=\mathcal{L}_{s-1}\{f(t)\}=F(s-1)$ (in case the notation was unfamiliar). I'm not sure how to continue with this. The factor of $e^{-t}$ seems to undo the shift to $s-1$, and I'm not sure if that's a problem or not.
I don't think I made a mistake with the $n$th derivative formula, but I think there might be a different way to express it that would make computation easier. Any help is appreciated!

REPLY [3 votes]: I don't think using the Rodrigues formula is convenient here. If one starts from the standard closed form representation of the Laguerre polynomials $$L_n(x)=\sum_{k=0}^n {{n} \choose {k}} (-1)^k \frac{x^k}{k!}$$ we have the rather straightforward calculation
$$\mathcal L_s(L_n(x)))=\int_0^\infty e^{-sx}L_n(x) dx=\sum_{k=0}^n {{n} \choose {k}}  \frac{(-1)^k}{k!} \int_0^\infty x^k e^{-sx}dx=\sum_{k=0}^n {{n} \choose {k}} (-1)^k s^{-k-1}= \\ s^{-1}\sum_{k=0}^n {{n} \choose {k}} (-s^{-1})^{k} =\left(\frac{s-1}{s} \right)^n \frac{1}{s} $$
having used the Gamma function relation $$\int_0^\infty x^k e^{-sx}dx= k! s^{-k-1}.$$