Question: Define the sequence $\{a\_{n}\}$ by $a\_{n} = n a\_{n-1} + (n+1)!$ for $n \geq 1$ and setting $a\_{0} = 1$. Solve this recursion completely.

I can solve this rather easily by an induction argument, where I found $a\_{n} = \frac{(n+1)!}{2}$. I was not, however, able to solve this using generating functions.

This is a problem from a past exam for one of my classes, and the professor is not fond of induction proofs - so I am thinking that another approach should be possible. If the method of generating functions is not practical, would there be a way to straightforward particular solution one can guess for the inhomogeneous relation?

Answer: 
**Hint:** Change $a\_n=n!b\_n$. You get $b\_n=b\_{n-1}+n+1$.
**Hint** Use *exponential* generating functions.