Finding explicit formula of $a_n$ It is given that $a_{n+1}=(n+1)a_n+2(n+1)!$ where $n>0$ and and $a_0=0$ , so find the explicit formula for $a_n$.
I tried to use exponential generating functions such that , lets say $A(x)= \sum_{n=0}^{\infty}a_n\frac{x^n}{n!}=(e^x-1)$ ,because $a_0=0$ .Then:

*

*$$a_{n+1}=\frac{n(A(x)-a_0)}{x}=\frac{nA(x)}{x}$$


*$$(n+1)a_n=nA(x)+A(x)$$


*$$2(n+1)!=2(n+1)\sum_{n=0}^{\infty}x^n=\frac{2n+2}{1-x}$$
As a result , i found that $$\frac{nA(x)}{x}=nA(x)+A(x)+\frac{2n+2}{1-x}$$
However , i could not take the $A(x)$ from the foregoing expression(i.e it does not provide an answer).I guess i am making a silly error but i stuck. So , can you help me to solve this recurrence relation using generating functions ?You do not have to follow my way , other methodds are also acceptable. Thanks in advance !!

Another way to find the formula is by solving
$$
a_n = a_n^h+a_n^p
$$
where $a_n^h$ is the solution for
$$
a_n^h = n a_{n-1}^h
$$
and $a_n^p$ is a convenient particular solution obeying
$$
a_n^h = n a_{n-1}^h+2n!
$$
For the homogeneous recurrence the solution is easily found as
$$
a_n^h = n! c_0
$$
The particular is solved proposing $a_n^p = c_0(n)n!$ and after substitution we get the recurrence
$$
c_0(n) = c_0(n-1) + 2
$$
with particular solution
$$
c_0(n) = 2n
$$
and finally we have
$$
a_n = n!c_0 + (2n)n! = (c_0+2n)n!
$$
but here $c_0 = a_0 = 0$ hence
$$
a_n = (2n)n!
$$
