Calculating the limit of a product Question: Let $a_1=1$ and $a_n=n(a_{n-1} +1)$ for $n=2,3, \ldots$
Define $P_n= (1+\frac{1}{a_1})(1+\frac{1}{a_2}) \cdots (1+\frac{1}{a_n}) ; \;\;\; n=1,2, \ldots$
Find $\lim_{n \to \infty} P_n$.
[I have outlined my (tedious) approach below, I'm merely curious whether there is a more elegant way to arrive at the result]
My Approach: I was able to arrive at the fact that $P_n=\frac{a_{n+1}}{(n+1)!}$ 
Then I expanded $a_{n+1}$ as follows
$$ a_{n+1}=(n+1)[a_n+1]$$
$$ = (n+1)[n(a_{n-1}+1)+1]$$
$$ = (n+1)(n)a_{n-1} + (n+1)(n)+(n+1)$$
$$ = (n+1)(n)[(n-1)\{a_{n-2}+1\}] + (n+1)(n)+(n+1)$$ 
$$ = (n+1)(n)(n-1)a_{n-2} + (n+1)(n)(n-1) + (n+1)(n)+(n+1)$$
Proceeding similarly, we arrive at the expansion 
$$ a_{n+1} = (n+1)!a_1 + (n+1)! + \frac{(n+1)!}{2!} + \frac{(n+1)!}{3!} + \cdots + \frac{(n+1)!}{n!}$$
and thus $$ P_n= \frac{a_{n+1}}{(n+1)!}= 1+ \frac{1}{1!} + \frac{1}{2!} +  \cdots + \frac{1}{n!}$$ 
$$ \lim_{n\to\infty} P_n = 1+ \frac{1}{1!} + \frac{1}{2!} + \cdots = e$$

Note that $$\frac{a_{n+1}}{(n+1)!}=\frac{a_n}{n!}+\frac{1}{n!}$$ this gives your final equation.
