TITLE: How to compute $\prod_{n=1}^\infty\left(1+\frac{1}{n!}\right)$?
QUESTION [40 upvotes]: Does
$$p=\prod_{n=1}^\infty\left(1+\frac{1}{n!}\right)$$
have any closed form in terms of known mathematical constants? The computer says 
  $$p=3.682154\dots$$
  but I don't even know how do devise the converging upper and lower bounds to obtain this result.


edit Jan. 15: I've got rid of the infinite product in favor of an fastly converging infinite sum over finite products here.

Thoughts:
$$p=\lim_{n\to \infty}p_n\hspace{.7cm}\text{where}\hspace{.7cm} p_n=p_{n-1}\cdot \left(1+\frac{1}{n!}\right)\hspace{.7cm}\text{with}\hspace{.7cm} p_1=2.$$
So I looked for an emerging pattern
$p_1=(1+\frac{1}{1!})$
$p_2=(1+\frac{1}{1!})(1+\frac{1}{2!})=(1+\frac{1}{1!}+\frac{1}{2!})+(\frac{1}{1!2!})$
$p_3=((1+\frac{1}{1!}+\frac{1}{2!})+\frac{1}{1!2!})(1+\frac{1}{3!})
=(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!})+(\frac{1}{1!2!}+\frac{1}{1!3!}+\frac{1}{2!3!})+\frac{1}{1!2!3!}$
It appears that 
$$p=1+\sum_{n=1}^\infty\sum_{m=1}^\infty a_{nm}$$
where $a_{1m}$ is the sum of terms with one inverse $\frac{1}{m!}$, and then $a_{2m}$ is the sum of (sums of) terms with two inverses $\frac{1}{r!s!}$. For example the term $\frac{1}{1!3!}$ is in the sum, and so I guess I need all the partitions into $n$ numbers. However, we don't want to count $\frac{1}{2!2!}$ and so it's more complicated. I guess the product can be written as a sum of term $(e-1)^n$ minus something, as for example
$(e-1)^2 
= \left(\frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+\cdots\right)\left(\frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+\cdots\right)
=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{  m!\,n!}$.
The logarithm of it is also a sum of sums which somewhat resembles the series expansion of the exponential function, but there, I think, the coefficients are powers of $\frac{1}{n!}$.

REPLY [4 votes]: I've been intrested in products aswell (see my question) the methode i used was this one, i hope it helps.
$$\prod_{i=b}^c 1+a_i$$=
$$1+\sum_{i=b}^{c} (a_i)+$$
$$1/2!((\sum_{i=b}^c (a_i))^2-\sum_{i=b}^c (a_i)^2$$
$$1/3!((\sum_{i=b}^c (a_i))^3-3\sum_{i=b}^c (a_i)^2\sum_{i=b}^c (a_i)+2\sum_{i=b}^c (a_i)^3)$$
$$1/4!((\sum_{i=b}^c (a_i))^4-6(\sum_{i=b}^c (a_i))^2\sum_{i=b}^c (a_i)^2+3(\sum_{i=b}^c (a_i)^2)^2+8(\sum_{i=b}^c (a_i)^3)\sum_{i=b}^c (a_i)-6\sum_{i=b}^c (a_i)^4)$$
These are the refined strirling numbers, but i guess you get the pattern, i dont master the skills yet to write this more efficient down.