How to show this, $$xL_n'(x) = nL_n(x)-nL_{n-1}(x)$$ Laguerre recursion relation from these two recursion relations, $$L'_{n+1}(x)-L'_n(x)+L_n(x)=0\\(n+1)L_{n+1}(x)-(2n+1-x)L_n(x)+nL_{n-1}(x)=0$$
Note that various books have various methods, I have looked up Arfken and Boas, they have two very different methods, but I want to find this from these two recursion relations.
First let us differentiate both sides of the second equation to get $$(n + 1)L'_{n + 1}(x) - (2n + 1)L'_n(x) + xL'_n(x) + L_n(x) + nL'_{n - 1}(x) = 0.$$ Then (suppressing the $(x)$ notation) we have that twice applying the first equation: \begin{align*} xL'_n & = (2n + 1)L'_{n} - (n + 1)L'_{n + 1} - L_n - nL'_{n - 1} \\ & = nL'_n + (n + 1)L'_n - (n + 1)L'_{n + 1} - (n + 1)L_n + nL_n - nL'_{n - 1} \\ & = nL'_n + nL_n - nL'_{n - 1} \\ & = nL'_n + nL_n - nL'_{n - 1} + nL_{n - 1} - nL_{n - 1} \\ & = nL_n - nL_{n - 1},\end{align*} the desired result.