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Let $a_1=1$ and $a_n=n(a_{n-1}+1)$ for $n=2,3,\ldots$

Find $$\lim_{n\to\infty}P_n\\ P_n = \left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)\left(1+\frac{1}{a_3}\right)\cdots\left(1+\frac{1}{a_n}\right); \ n=1,2,\ldots$$

My approach:

$$ P_n = \left(\frac{a_1+1}{a_1}\right)\left(\frac{a_2+1}{a_2}\right)\cdots\left(\frac{a_n+1}{a_n}\right)$$

or $$P_n = \left(\frac{a_2}{2a_1}\right)\left(\frac{a_3}{3a_2}\right)\left(\frac{a_4}{4a_3}\right)\cdots\left(\frac{a_{n+1}}{(n+1)a_n}\right)$$

or $$P_n = \left(\frac{a_{n+1}}{a_1(n+1)!}\right)$$

or $$\lim_{n\to\infty} P_n =\lim_{n\to\infty} \left(\frac{a_{n+1}}{(n+1)!}\right)$$

How to find $\lim_\limits{n\to\infty}a_{n+1}$? I am stuck here.

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Hint. From $$ a_n=n(a_{n-1}+1), $$ by dividing by $n!$ one gets $$ \frac{a_n}{n!}=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!} $$or $$ \frac{a_n}{n!}-\frac{a_{n-1}}{(n-1)!}=\frac{1}{(n-1)!} $$ then by summing from $n=2$ to $n=N$ terms telescope giving $$ \frac{a_N}{N!}-\frac{a_{1}}{1!}=\sum_{n=2}^N\frac{1}{(n-1)!} $$ that is $$ \frac{a_N}{N!}=\sum_{n=0}^{N-1}\frac{1}{n!},\quad N\ge1. $$ One may recall that

$$ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},\quad x\in \mathbb{R}. $$

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