Factorize $x+{1\over 2}x^2+{1\over 3}x^3+ \cdot\cdot$ by $1-x$ I want to derive explicit formula for given recursive relation below:
$$a_{n+1} = (n + 1)a_n + n!$$ 
for $n ≥ 0$ and $a_0 = 0$.
I had exploited $EGF$, resulting in:
$$g(x)\cdot(1-x) = x+{1\over 2}x^2+{1\over 3}x^3+ \cdot\cdot$$
Thus to derive the explicit formula of $a_n$, I am thinking about how I can manage the $RHS$ to be factorized by $1-x$ so that I can have $a_n$ for corresponding $x^n/n!$
Any advice to proceed further? 
I yet haven't took the abstract algebra class where I seemingly guess I could have more chance to be familiar to polynomial series.

We can obtain the numbers $a_n (n\geq 1, a_0=0)$ by calculating the coefficients of the exponential generating function
\begin{align*}
g(x)=a_1x+a_2\frac{x^2}{2!}+a_3\frac{x^3}{3!}+a_4\frac{x^4}{4!}+\cdots
\end{align*}

We obtain
  \begin{align*}
\color{blue}{g(x)}&=\left(x+\frac{1}{2}x^2+\frac{1}{3}x^3+\frac{1}{4}x^4+\cdots\right)\frac{1}{1-x}\\
&=\left(x+\frac{1}{2}x^2+\frac{1}{3}x^3+\frac{1}{4}x^4+\cdots\right)\left(1+x+x^2+x^3+x^4+\cdots\right)\tag{1}\\
&=x+\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3
+\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)x^4+\cdots\tag{2}\\
&=\sum_{n=1}^\infty H_nx^n\tag{3}\\
&=\sum_{n=1}^\infty\color{blue}{n!H_n}\frac{x^n}{n!}\tag{4}
\end{align*}
We conclude the numbers are $$\color{blue}{a_n=n!H_n  \qquad n\geq 1}$$.

Comment:


*

*In (1) we use the geometric series expansion.

*In (2) we multiply the series of the right-hand side and collect the terms with equal powers of $x$. We observe the coefficients are 
\begin{align*}
1,\,1+\frac{1}{2},\,1+\frac{1}{2}+\frac{1}{3},\,1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4},\cdots
\end{align*}
which are  the harmonic numbers denoted with $H_n$.

*In (3) we use the sigma notation for brevity.

*In (4) we write the series as exponential generating series to better see the coefficients $a_n$.

Note: If we write the recursion for $a_n$ as
  \begin{align*}
\frac{a_{n+1}}{(n+1)!}&=\frac{a_n}{n!}+\frac{1}{n+1}\qquad\qquad n\geq 1\\
a_0&=0
\end{align*}
  then the solution $a_n=n!H_n$ might be seen easily.

