From Wikipedia, the Laguerre Differential Equation is defined as follows: \begin{align} x y'' + (v + 1 - x) y' + \lambda y = 0 \end{align} By definition, the solution of this differential equation is expressed as : \begin{align} \, L_{\lambda}^v (x)=\frac{d^v}{dx^v}[L_\lambda (x)] \\ \end{align}
Let's do a power series method. Then let the arbitrary constant set to '1' and the solution to be a finite polynomial which depends on $ \lambda $ and some “clever” algebra tricks, someone can obtain the Rodriguez formula for the solution of the general Laguerre differential equation: \begin{align} x y'' + (1 - x) y' + \lambda y = 0 \end{align}
where, \begin{align} y = L_{\lambda}= \frac{e^x}{{\lambda}!} \frac{d^{\lambda}}{dx^{\lambda}} [x^n e^{-x}]\end{align}
Now, differentiating the Laguerre equation 'v' times we get into the format:
\begin{align} x f'' + (v + 1 - x) f' + ( \lambda - v)y = 0 \end{align} \begin{align} f=\frac{d^v}{dx^v}[L_\lambda (x)] \end{align}
My question is, as the formula suggests, by taking 'v' differentiation should bring our general Laguerre differential equation to the associated Laguerre differential equation but apparently it doesn't. Then why it is said to be like this? Or is there more formulation need to be done after taking differentiation 'v' times to show that the following formula is true?