How to prove that the recurrence $a_{n}=a_{n-1}+n^2a_{n-2}$ gives $(n+1)!$ without induction 
Define the sequence $\{a_n\}$ by $a_{1}=2,a_{2}=6$, and for $n>2$, 
  $$a_{n}=a_{n-1}+n^2a_{n-2}$$

show that
$$a_{n}=(n+1)!$$
I know if we use induction,it is easy to prove it.
$$n!+n^2(n-1)!=n![1+n]=(n+1)!$$
But without using induction, can we prove this result?
 A: Divide
$$
a_n=a_{n-1}+n^2 a_{n-2}
$$
by $a_{n-1}$:
$$
\frac{a_n}{a_{n-1}}=1+n^2\frac{a_{n-2}}{a_{n-1}}
$$
Let $x_n=\frac{a_n}{a_{n-1}}$; so,
$$
x_n=1+\frac{n^2}{x_{n-1}}.
$$
$x_n=n+1$ is a solution of it:
$$
n+1=1+\frac{n^2}{(n-1)+1}.
$$
So, $a_n/a_{n-1}=n+1$ and $a_n=C(n+1)!$. $a_1=2\Longrightarrow C=1$.
A: Well, if we know that
the solution to
$a_{n}=a_{n-1}+n^2a_{n-2}
$
is
$a_n
=(n+1)!
$,
let's let
$a_n
=(n+1)!b_n
$
(with initial values
$b_1 = b_2 = 1$
and see what happens.
$(n+1)!b_n
=n!b_{n-1}+n^2(n-1)!b_{n-2}
=n!b_{n-1}+nn!b_{n-2}
$
or
$(n+1)b_n
=b_{n-1}+nb_{n-2}
$.
And for this,
we see that
$b_n = 1$
is the solution.
Note that from
$(n+1)b_n
=b_{n-1}+nb_{n-2}
$,
$(n+1)b_n-(n+1)b_{n-1}
=-nb_{n-1}+nb_{n-2}
$
or
$(n+1)(b_n-b_{n-1})
=-n(b_{n-1}-b_{n-2})
$.
Since
$b_1 = b_2 = 1$,
this implies that
$b_n = 1$ for all $n$.
A: Consider the sequence $a_{1}=2,a_{2}=6$, $a_{n}=a_{n-1}+n^2a_{n-2}$. Let $a_{n} = \Gamma(n+2) \, b_{n}$ for which it is seen that:
\begin{align}
\Gamma(n+2) \, b_{n} = \Gamma(n+1) \, b_{n-1} + n \, \Gamma(n+1) \, b_{n} 
\end{align}
or
\begin{align}
(n+1) \, b_{n} = b_{n-1} + n \, b_{n-2}
\end{align}
where $b_{1} = b_{2} = 1$. By calculating the next set of terms it is quickly realized that $b_{n} = 1$ for $n \geq 1$. Utilizing this result it is determined that the solution to $a_{n}=a_{n-1}+n^2a_{n-2}$ is $a_{n} = \Gamma(n+2)$.  
