solution of recursive equation $a_{n} + na_{n-1} = n!$

Question

Can anyone solve this equation and give a general formula for $a_{n}$. The initial value is $a_{0} = 1$ and $n \geq 1$.
$$a_{n} + na_{n-1} = n!$$

Answer 1

Hint: Set $b_n:=\frac{a_n}{n!}$.

Answer 2

HINT:

$$a_0=1\\a_1+1a_0=1!\implies a_1=1-1=\color{red}0\\a_2+2a_1=2!\implies a_2=\color{blue}{2!}\\a_3+3a_2=3!\implies a_3=3!-3(2!)=\color{red}0\\a_4+4a_3=4!\implies a_4=\color{blue}{4!}\\\cdots$$

Answer 3

By dividing both sides of the equation to $n!$ we have:
$$\frac{a_{n}}{n!} + \frac{a_{n-1}}{(n-1)!} = 1$$
Assuming $b_{n} = \frac{a_{n}}{n!}$
$$b_{n} + b_{n-1} = 1$$