Let $a_{1} = 1$ and $a_{n} = n(a_{n-1}+1)\;\forall n\geq 2.$ Then $ \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(1+\frac{1}{a_{r}}\right)$ 
Let $a_{1} = 1$ and $a_{n} = n(a_{n-1}+1)\;\forall n\geq 2.$ Then $\displaystyle \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(1+\frac{1}{a_{r}}\right)$

$\bf{My\; Try::}$ Given  $a_{n} = n(a_{n-1}+1)$
Replace $n\rightarrow n+1\;,$ We get $$a_{n+1}=(n+1)(a_{n}+1)\Rightarrow \frac{a_{n+1}}{n+1} = a_{n}+1$$
Now how can i solve it, Help required, Thanks
 A: Note that $$a_{r+1}=(r+1)(a_r+1) \implies \frac{a_{r+1}}{r+1}=a_r+1$$
Now $$\displaystyle \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(1+\frac{1}{a_{r}}\right)$$
$$=\displaystyle \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(\frac{a_r+1}{a_{r}}\right)$$
$$=\displaystyle \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(\frac{1}{a_{r}}\cdot \frac{a_{r+1}}{r+1}\right)$$
$$=\displaystyle \lim_{n\rightarrow \infty} \prod^{n}_{r=1}\left(\frac{1}{r+1}\cdot \frac{a_{r+1}}{a_{r}}\right)$$
$$=\displaystyle \left(\frac{1}{1+1}\cdot \frac{a_{1+1}}{a_{1}}\right)\left(\frac{1}{2+1}\cdot \frac{a_{2+1}}{a_{2}}\right)\left(\frac{1}{3+1}\cdot \frac{a_{3+1}}{a_{3}}\right)\ldots \left(\frac{1}{r+1}\cdot \frac{a_{r+1}}{a_{r}}\right)\ldots $$
$$=\displaystyle \left(\frac{1}{2}\cdot \frac{\not a_{2}}{a_{1}}\right)\left(\frac{1}{3}\cdot \frac{\not a_{3}}{\not a_{2}}\right)\left(\frac{1}{4}\cdot \frac{\not a_{4}}{\not a_{3}}\right)\ldots \left(\frac{1}{r+1}\cdot \frac{\not a_{r+1}}{\not a_{r}}\right)\ldots $$
Hope, this will serve as a sufficient hint since you wanted a hint only.
A: Note that
$$
\frac{a_n}{n!}=\frac{a_{n-1}}{(n-1)!}+\frac1{(n-1)!}=1+1+\frac1{2!}+…+\frac1{(n-1)!}
$$
and as SchrodingersCat demonstrated, the finite products are
$$
\prod_{r=1}^n\left(1+\frac1{a_r}\right)=\frac{a_{n+1}}{(n+1)!}
$$
which now has an obvious limit.
