Let $a_1 = 1$ and $a_n = n(a_{n-1}+1)$ for $n=2,3,..$. Then $$\lim_{n\to\infty}P_n:=\lim_{n\to\infty}(1+1/a_1)(1+1/a_2)\cdot\ldots\cdot(1+1/a_n)$$ is which of the following:
a) $1+e$
b) $e$
c) $1$
d) $\infty$
I tried to simplify it as such, but got stuck.
$$ \frac{a_n}{n!} = \frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!} $$ $$ =\frac{1}{(n-1)!}+\frac{1}{(n-2)!}+\frac{a_{n-2}}{(n-2)!} $$ $$ =\frac{1}{(n-1)!}+\frac{1}{(n-2)!}+\cdots $$ $$ =\sum_{k=0}^{n-1}\frac{1}{k!} $$
if $na_n = (a_{n-1}+1)\Rightarrow n!(a_1\cdot ...\cdot a_n) = (a_{n-1}+1)\cdot ...\cdot (a_{1}+1)\Rightarrow P_{n-1} = n!a_n$. By the expression that you calculated: $P_n = (n!)^2\sum_{k=0}^{n-1}\frac{1}{k!}$ and you can use that $\sum_{k=0}^{n-1}\frac{1}{k!}\to e$ when $n\to\infty$ then $P_n\to\infty$