= Solution
Differentiating $v=x^ne^{-x}$ gives $xv'=(n-x)v$. Applying the <Leibniz rule> $n+1$ times to this identity gives
$$
xv^{(n+2)}+(n+1)v^{(n+1)}=(n-x)v^{(n+1)}-(n+1)v^{(n)}.
$$
Only two terms survive on each side because the multiplying factors are linear. Rearranging yields $xv^{(n+2)}+(x+1)v^{(n+1)}+(n+1)v^{(n)}=0$. Since $v^{(n)}=n!e^{-x}L_n$, the next two <derivatives> are $n!e^{-x}(L_n'-L_n)$ and $n!e^{-x}(L_n''-2L_n'+L_n)$. Substituting and cancelling $n!e^{-x}$ gives
$$
\boxed{xL_n''+(1-x)L_n'+nL_n=0}.
$$
This is <deriving the Laguerre equation from Rodrigues' formula>, exhibiting $L_n$ as an <eigenfunction> of the <Laguerre differential operator on polynomials>.
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