Finding $\lim _{n\to\infty}(1+\frac1{a_{1}})(1+\frac1{a_{2}})\cdots(1+\frac1{a_{n}})$, where $a_1=1$ and $a_n=n(a_{n-1}+1)$ 
Let $a_{1}=1$ and $a_{n}=n\left(a_{n-1}+1\right)$ for $n=2, 3, \ldots$. Define
$$
P_{n}=\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right)
$$
for $n=1, 2, \ldots$. Find $\lim _{n \rightarrow \infty} P_{n}$.

My work:
Using the expression I am getting
$$a_{2} = 4$$
$$a_{3} = 15$$
$$a_{4} = 64$$
What to do next?
Edit 1. After taking log, I am getting the final series as $\log(1+1/n^{2})$. So, the answer should be $0$.
 A: Use the fact that:
$$
1+\frac{1}{a_{k}} =
\frac{a_{k} + 1}{a_{k}} = 
\frac{a_{k} + 1}{k(a_{k-1} + 1)}
$$
This will allow us to express $P_n$ as $\frac{a_{n+1}}{(n+1)!}$. Now, expand it using the recurrence relation of the sequence $\left(a_n\right)$:
\begin{align}
\frac{a_{n+1}}{(n+1)!} &= 
\frac{a_{n}}{n!} + \frac{1}{n!} \\&= 
\frac{a_{n-1}}{(n-1)!} + \frac{1}{(n-1)!} + \frac{1}{n!} \\&=
\frac{a_{n-2}}{(n-2)!} + \frac{1}{(n-2)!} + \frac{1}{(n-1)!} + \frac{1}{n!} \\&= 
\cdots \\&= 
\frac{a_1}{1!} + \sum^{n}_{k=1} \frac{1}{k!}
\end{align}
A: $$a_{n}=n\left(a_{n-1}+1\right)\quad\text{with} \quad a_1=1 \implies a_n=e n \Gamma (n,1)$$
$$P_n=\prod_{k=1}^n \left(1+\frac{1}{a_k}\right)= \prod_{k=1}^n \left(1+\frac{1}e\frac{1}{ k \,\Gamma (k,1)}\right)$$ which generate the sequence
$$\left\{2,\frac{5}{2},\frac{8}{3},\frac{65}{24},\frac{163}{60},\frac{1957}{720},\frac
   {685}{252},\frac{109601}{40320},\frac{98641}{36288},\frac{9864101}{3628800},\cdots\right\}$$
These fractions correspond to the numerators and denominators of $\sum_{k=0}^n \frac 1 {k!}$ (sequences $A061354$ and $A061355$ in $OEIS$) .
So, it seems that when $n\to \infty$, $P_n$ tends to ... a very well knonw number.
