<s>[INST] How do I get the closed form of this recurrence using generating functions? Recurrence: $T(n) = n + nT(n-1)$ and $T(0) = 0$.
What I tried:
Let $G(x) = \sum_{n=0}^{\infty} T(n) x^n$ so that $xG'(x) = \sum_{n=0}^{\infty} nT(n)x^n$
Solving:
\begin{align}
G(x) &= \sum_{n=0}^{\infty} T(n) x^n 
\\&= 0x^0 + \sum_{n=1}^{\infty} T(n) x^n 
\\&= \sum_{n=1}^{\infty} n x^n + \sum_{n=1}^{\infty} nT(n-1) x^n 
\\&= -0x^0 + \sum_{n=0}^{\infty} n x^n + x\sum_{n=1}^{\infty} nT(n-1) x^{n-1} 
\\&= \frac{x}{(1-x)^2} + x\sum_{n=0}^{\infty} (n+1)T(n) x^{n} 
\\&= \frac{x}{(1-x)^2} + x\sum_{n=0}^{\infty} nT(n) x^{n} + x\sum_{n=0}^{\infty} T(n) x^{n} 
\\&= \frac{x}{(1-x)^2} + x^2G'(x) + xG(x) 
\end{align}
And $G(x) = \frac{x}{(1-x)^2} + x^2G'(x) + xG(x)$ rearranges to $G(x) = \frac{x}{(1-x)^3} + \frac{x^2}{1-x}G'(x)$
No idea where to take it from here.
 [/INST] $$G(x)-xG'(x)=\frac{x}{(1-x)^2}$$
Now use the fact that 
$$\left(\frac{G(x)}{x} \right)'=\frac{G'(x) x -G(x)}{x^2}$$
to reduce your equation to 
$$\left(\frac{G(x)}{x} \right)'=-\frac{1}{x (1-x)^2}$$
Integrate and you are done.
Edit To address the updated question:
$$  G'(x) -\frac{1-x}{x^2}G(x) =-\frac{1}{x(1-x)^2} $$
is a first order linear differential equation. 
You can solve it by using the method listed in these notes:
 Linear Differential Equations
</s>