Find the value of this infinite product $\prod_{n=1}^{\infty }\left [ 1+\frac{1}{a_{n}} \right ]$? Where $a_{1}=1$ and $a_{n+1}= \left ( n+1 \right )\left ( 1+a_{n} \right )$, 
$\forall \left ( n\epsilon \mathbb{N} \right )$
I have found that for this infinite product to be convergent, the series $\sum \ln \left ( 1+\frac{1}{a_{n}} \right)$ must also be convergent, and that if it has a finite sum, let's name it s, the value of the infinite product  is $ e^{s}$.
I am not sure whether $a_{n}=n\cdot \left ( 1+a_{n-1} \right )$ also holds true.
The given series passes the n-th term test, ok but how do I apply another test to it?

Note that 
$$a_{n+1}=(n+1)(1+a_n)\Rightarrow \frac{a_{n+1}}{1+a_{n}}=n+1$$
Consider the finite product:
$$\mathcal{P}_n=\prod_{n=1}^{N}(1+\frac{1}{a_n})=\prod_{n=1}^{N}\frac{1+a_n}{a_n}=\prod_{n=1}^{N}\frac{a_{n+1}}{(n+1)a_n}=\prod_{n=1}^{N}\frac{1}{(n+1)}\cdot \prod_{n=1}^{N}\frac{a_{n+1}}{a_n}=\frac{1}{N!}\cdot\frac{a_N}{a_1}=\frac{a_N}{N!}$$
Now we need to find an expression for the general term $a_n$. Notice that 
$$a_{n+1}=(n+1)(1+a_n)=(n+1)+(n+1)n(1+a_{n-1})=(n+1)+(n+1)n+(n+1)n(n-1)(1+a_{n-2})=...=(n+1)+(n+1)n+(n+1)n(n-1)+..+(n+1)n(n-1)(n-2)...2\cdot(1+1)$$
In other words
$$a_{n+1}=\frac{(n+1)!}{n!}+\frac{(n+1)!}{(n-1)!}+..+2\cdot\frac{(n+1)!}{1!}$$
Hence 
$$a_N=\frac{N!}{(N-1)!}+\frac{N!}{(N-2)!}+..+2\cdot\frac{N!}{1!}$$
for all $N\geq1$ so 
$$\mathcal{P}_N=\frac{a_N}{N!}=\frac{1}{(N-1)!}+\frac{1}{(N-2)!}+..+\frac{2}{1!}$$
One can see that 
$$1\leq \mathcal{P}_N<\frac{1}{(N-1)(N-2)}+\frac{1}{(N-2)(N-3)}+..+2=3-\frac{1}{N-1}$$
then
$$1\leq\lim_{N\to\infty}\mathcal{P}_N\leq3$$
Indeed the limit is $e$. Check it with the Taylor expansion of $e^x$ at $x=1$.
