Limit of a sequence $a_1=1;a_{n+1}=(n+1)(1+a_n)$ Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$
If $P_n=\prod_1^n(1+a_i^{-1})$
then $$\lim_{n\to\infty}P_n $$is ?



I did:
$$P_n=\prod_1^n\frac{(1+a_i)}{a_i}
=\frac{a_{n+1}}{a_1}\prod1^{n-1}\frac{(1+a_i)}{a_{i+1}}
=\frac{a_{n+1}}{a_1}\prod_1^{n-1}\frac1{i+1}=\frac{a_{n+1}}{n!}$$
Now how do I find the limit?
$a_n = n + na_{n-1}$. This is $a_n = n + n(n-1) + n(n-1)(n-2) + \ldots + n!$. In turn, $\frac {a_n}{n!} = \frac{1}{(n-1)!} + \frac{1}{(n-2)!} + \frac{1}{(n-3)!} + \ldots + 1$, which approaches the Taylor series for $e$.
Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$
We have $$P_n=\prod_{i=1}^n(1+a_i^{-1})$$
then 

$$
P_n
=\prod_{i=1}^n (1 + \frac{1}{a_i})
=\left (1 + \frac{1}{n(1+a_{n-1})}\right )\prod_{i=1}^{n-1} (1+a_i^{-1}  )
=\left (\frac{n(1+a_{n-1}) + 1}{n(1+a_{n-1})}\right )\prod_{i=1}^{n-1} (1+a_i^{-1}  )
=\left (\frac{n(1+a_{n-1}) + 1}{n(1+a_{n-1})}\right )
 \left (\frac{(n-1)(1+a_{n-2}) + 1}{(n-1)(1+a_{n-2})}\right )\prod_{i=1}^{n-2} (1+a_i^{-1}  )
= \ldots 
\approx \frac{n! + \prod_{i=1}^n a_i}{n! + \prod_{i=1}^n a_i} = 1
$$