Let $a_1=1$ and $a_n=n(a_{n-1}+1)$ for $n=2,3,...$. Let $a_1=1$ and $a_n=n(a_{n-1}+1)$ for $n=2,3,...$. Define $$P_n=\left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)...\left(1+\frac{1}{a_n}\right).$$ Find $\lim\limits_{n\to\infty} P_n$. 
My approach: $$P_n=\left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)...\left(1+\frac{1}{a_n}\right)$$ $$=\left\{\frac{(a_1+1)(a_2+1)...(a_n+1)}{a_1.a_2...a_n}\right\}$$ $$=\left\{\frac{1}{a_1}\left(\frac{a_1+1}{a_2}\right)\left(\frac{a_2+1}{a_3}\right)...\left(\frac{a_{n-1}+1}{a_n}\right)(1+a_n)\right\}$$ $$=\left\{\frac{1}{a_1}.\frac{1}{2}.\frac{1}{3}.....\frac{1}{n}(1+a_n)\right\}$$ $$=\frac{1+a_n}{n!}, \forall n\in\mathbb{N}.$$ 
Thus, we need to find $$\lim_{n\to\infty} \frac{1+a_n}{n!}.$$ How to proceed after this?

$$\begin{align}
a_{n}&=na_{n-1}+n\\
&=n[(n-1)a_{n-2}+(n-1)]+n\\
&=n(n-1)[(n-2)a_{n-3} + (n-2)]+n(n-1)+n\\
& \phantom{a bit of space here would be nice}\vdots\\
&=\left[n(n-1)(n-2)\dots 3 \cdot 2\cdot 1\right]\left( 1+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{(n-1)!} \right)\\
&=n!\left(1+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{(n-1)!}\right)\rightarrow n!e, \quad (n \to \infty)
\end{align}
$$
