How do I solve $a_{n+1}=(n+1)a_n$ by recurrence relations given $a_0=1$? I realize that it equals $n!$ by inspection, but how do I solve it as a first-order non-homogenous recurrence relation (with characteristic polynomials like in linear algebra)? I have tried to solve it like this:
$a_{n+1}=(n+1)a_n$
$a_{n}=(n)a_{n-1}$
But it is not like adding a constant, in which I can just do this and subtract. Dividing is of no help either. How should I approach such a problem?
P. S. I have not learned differential equations, and don't wish to do algebra/function bash as in Solving this non-homogenous recurrence relation. This answer Solving this non-homogenous recurrence relation is too vague for me, and I don't know how to solve the equation using the method.
Here is an approach. Note that $$ \begin{split} a_n &= n a_{n-1} = n \left( (n-1) a_{n-2} \right) \\ &= n (n-1) a_{n-2} \\ &= n (n-1) (n-3) a_{n-2} \\ & \ldots\\ &= n \times \ldots \times (n - (n-1)) a_{n-(n-1)}\\ &= a_1 \left(\prod_{k=1}^n k \right) \\ &= a_1 \times n! \end{split} $$ and it's easy to show from the recurrence that since $a_0 = 1$ then $a_1 = 1$ as well.
Another approach is to substitute $a_n = b_n \cdot n!$ to get the recurrence $$ b_{n+1} \cdot (n+1)! = a_{n+1} = (n+1) a_n = (n+1) \cdot n! \cdot b_n $$ which is equivalent to $b_{n+1} = b_n$ which you can solve with usual methods to get $b_n \equiv 1$.
Last one, if you insist on doing this specifically using the characteristic polynomials, is to take logs in some convenient base, and let $b_n = \log (a_n)$ obtaining the recurrence $$ b_{n+1} = b_n + \log (n+1) $$ which you can try to solve using the usual methods.