Derive Rodrigues’ formula for Laguerre polynomials 
Derive Rodrigues’ formula for Laguerre polynomials
$$
L_n(x)=\frac{e^x}{n!}.\frac{d^n}{dx^n}(x^ne^{-x})
$$

The Rodrigues’ formula for Hermite polynomials  can be obtained by taking $n^{th}$ order partial derivatives of its generatig function
$$
g(x,t)=\sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}=1+tH_1(x)+\frac{t^2}{2!}H_2(x)+\cdots\cdots\cdots+\frac{t^n}{n!}H_n(x)+\cdots\cdots\cdots\\
\frac{\partial^n}{\partial t^n}\Big(e^{2xt-t^2}\Big)=H_n(x)+\frac{(n+1)n(n-1)\cdots2}{(n+1)!}tH_{n+1}(x)+\cdots\\
H_n(x)=\Bigg[\frac{\partial^n}{\partial t^n}\Big(e^{2xt-t^2}\Big)\Bigg]_{t=0}=e^{x^2}\Bigg[\frac{\partial^n}{\partial t^n}\Big(e^{-(x-t)^2}\Big)\Bigg]_{t=0}\\
=(-1)^ne^{x^2}\Bigg[\frac{\partial^n}{\partial x^n}\Big(e^{-(x-t)^2}\Big)\Bigg]_{t=0}=(-1)^ne^{x^2}\Bigg[\frac{\partial^n}{\partial x^n}\Big(e^{-x^2}\Big)\Bigg]
$$
Generating function for laguerre polynomials is $g(x,t)=\dfrac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^\infty L_n(x) t^n$
I do not think the same technique applies in the case of Laguerre polynomials. So how do I derive that for Laguerre polynomials ?
 A: Note that the Taylor-Maclaurin formula cannot be used directly (as in the case of Hermite polynomials) as there is some $n$ dependency inside the $n^{th}$ derivative. How can we get around this ? ... Let use a coefficient extrator
\begin{eqnarray*}
x^n= [u^0]: \frac{u^{-n}}{1-xu}.
\end{eqnarray*}
We have
\begin{eqnarray*}
\sum_{n=0}^{\infty} L_n(x) t^n &=&[u^0]: e^x \sum_{n=0}^{\infty} \left(\frac{t}{u} \right)^n \frac{1}{n!}  \frac{d^n}{dx^n} \frac{e^{-x}}{1-xu} \\
&=&[u^0]: e^x  \frac{e^{-(x+t/u)}}{1-u(x+t/u)} \\
&=& \frac{1}{1-t} [u^0]:  \frac{e^{-t/u}}{1-\frac{ux}{1-t}}. \\
\end{eqnarray*}
Expand these two functions and observe that the central term is ... what we want
\begin{eqnarray*}
\left( \sum_{i=0}^{\infty} \frac{(-t/u)^i}{i!} \right) \left( \sum_{j=0}^{\infty} \left( \frac{ux}{1-t} \right)^j \right) = 
\cdots+ \sum_{k=0}^{\infty} \frac{1}{k!} \left( \frac{-xt}{1-t} \right)^k + \cdots 
\end{eqnarray*}
Now reverse engineer all this ... and your result follows.
A: If you are allowed to use the explicit representation for Laguerre polynomials,
$$ L_n(x) =\sum_{k=0}^n \binom{n}{k} \frac{(-x)^k}{k!}  \quad ,$$
then the result follows easily from the Leibniz differentiation rule for products
$$ (f(x) \ g(x) )^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x) .$$
Refer to the Rodrigues formula you want to prove. Let $f(x)=x^n$ and $g(x)=e^{-x}.$ Then $f^{(k)}(x) = x^{n-k}\ n!/(n-k)! $ and
$g^{(n-k)}(x) = (-1)^{n-k}e^{-x}.$  Now clean it up and redo the sum in reverse order, i.e., $k \to n-k,$ and you'll get the top formula.
