Laguerre polynomials has the recurrence relation $$(n+1) L_{n+1} (x)=(2n+1-x) L_n (x)-nL_{n-1} (x)$$
In proving this, I differentiate the generating function of laguerre polynomials $$g(x,t)=\frac{e^{-\frac{xt}{1-t}}}{(1-t)}=\sum_{n=0}^\infty L_n (x)t^n $$
The process goes like this: \begin{align} \frac{\partial g}{\partial t} &=\frac{(1-t) D_t \left(e^{-\frac{xt}{1-t}}\right)-e^{-\frac{xt}{1-t}} D_t \left(1-t\right) }{(1-t)^2}\\ &=e^{-\frac{xt}{1-t}} \frac{1-t-x}{(1-t)^3} \end{align} $$(1-t)^2 \frac{\partial g}{\partial t}=(1-t-x)g \tag 1$$
solving the right-hand side of eq. (1), we have: \begin{align} (1-t-x)g &=(1-t-x) \sum_{n=0}^\infty L_n (x)t^n\\ &=L_0 (x)-xL_0 (x)+\sum_{n=1}^\infty(L_n (x)-L_{n-1} (x)-xL_n (x) ) t^n \end{align}
also, Solving the left-hand side of eq. (1), we have: $$ \begin{align} (1-t)^2 \sum_{n=0}^\infty nL_n (x) t^{n-1}&=(1-2t+t^2)\sum_{n=0}^\infty nL_n (x) t^{n-1}\\ &=\sum_{n=0}^\infty nL_n (x)(t^{n-1}-2t^n+t^{n+1})\\ &=L_1 (x)+\sum_{n=1}^\infty t^n [(n+1) L_{n+1}(x)-2nL_n (x)+(n-1)L_{n-1} (x) ]. \end{align} $$
My question is: How did the derivative of the generating function with respect to $t$ in the left hand side equal to $\sum_{n=0}^\infty nL_n (x) t^{n-1}$ that is, $$ \frac{\partial g}{\partial t}=\sum_{n=0}^\infty nL_n (x) t^{n-1} $$ Thanks! :)