Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})\cdots(1+\frac{1}{a_n})$, where $a_1=1$, $a_n=n(1+a_{n-1})$ 
Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\cdots(1+\frac{1}{a_n})$,
  where $a_1 = 1$, $a_n = n(1+a_{n-1})$

\begin{align*}
&\lim_{n\to\infty} \left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)\cdots\left(1+\frac{1}{a_n}\right) \\
&= \lim_{n\to\infty} \frac{1}{a_1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdots \frac{1}{n} \cdot (a_n + 1) \\
&= \lim_{n\to\infty} \frac{a_n + 1}{n!}.
\end{align*}
Then I'm stuck. How to proceed?

Observe that
$$a_n=n+na_{n-1}$$
$$a_n=n+n(n-1)+n(n-1)a_{n-2}$$
and so on, giving us eventually
$$a_n=\sum_{k=1}^n\frac{n!}{(k-1)!}$$
Now we have
$$\lim_{n\to\infty}\frac{1}{n!}(a_n+1)=\lim_{n\to\infty}\frac{a_n}{n!}$$
$$=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{(k-1)!}$$
$$=\sum_{k=0}^\infty\frac{1}{k!}=e$$
