How to compute $\prod_{n=1}^\infty\left(1+\frac{1}{n!}\right)$? 
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 $\frac{1}{n!}$.
 A: Note:  Steven Stadnicki made a good point
about the not-so-good convergence 
of my suggested computationa; method.
I am modifying this to try to correct this.
Expanding Samrat's observation,
for any positive integer $m$,
$\begin{align}
\ln p
&=\sum_{n=1}^{\infty}\ln \left(1+\frac{1}{n!}\right)\\
&=\sum_{n=1}^{m}\ln \left(1+\frac{1}{n!}\right)+\sum_{n=m+1}^{\infty}\ln \left(1+\frac{1}{n!}\right)\\
&= G_m+F_m\\
\text{where}\\
F_m &=\sum_{n=m+1}^{\infty}\ln \left(1+\frac{1}{n!}\right)\\
&=\sum_{n=m+1}^{\infty}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k(n!)^k}\\
&=\sum_{k=1}^{\infty}\sum_{n=m+1}^{\infty} \frac{(-1)^{k-1}}{k(n!)^k}\\
&=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=m+1}^{\infty} \frac{1}{(n!)^k}\\
\end{align}
$
Let $f_m(k) = \sum_{n=m+1}^{\infty} \frac{1}{(n!)^k}
$,
so
$F_m = \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} f_m(k)
$.
$f_m(1)$ is $e$ 
minus the start of its series,
and so is transcendental.
$f_m(2)$
is $I_0(2)$
minus the start of its series,
where $I_0(x)$
is the modified Bessel function of the first kind.
I do not know if $I_0(2)$
is transcendendal,
but $J_0(1)$ is known to be,
so I would be willing to bet that
$all$ the $f_m(k)$
are transcendental.
I will now get an upper bound
on  $f_m(k)$
to aid in the computation of $F_m$.
$f_m(k) 
= \sum_{n=m+1}^{\infty} \frac{1}{(n!)^k}
= \frac{1}{(m!)^k}\sum_{n=m+1}^{\infty} \left(\frac{m!}{n!}\right)^k
$.
If $n > m$,
$\frac{n!}{m!}
=\prod_{k=1}^{n-m} (m+k)
\ge (m+1)^{n-m}
$,
so
$f_m(k) 
\le \frac{1}{(m!)^k}\sum_{n=m+1}^{\infty} \left(\frac{1}{(m+1)^{n-m}}\right)^k
= \frac{1}{(m!)^k}\sum_{n=1}^{\infty} \frac{1}{(m+1)^{nk}}
= \frac{1}{(m!)^k}\frac{(m+1)^{-k}}{1-(m+1)^{-k}}
= \frac{1}{(m!)^k((m+1)^k-1)}
$.
Since the $f_m(k)$ are decreasing,
$|F_m| < f_m(1) 
< \frac{1}{m!m}
$.
This means that the error
in using the first $m$ terms in the product
is less than this bound.
This somehow seems
a more obvious result than I would like,
but I will have to leave it at this
since I do not see a more effective way of
estimating the sum.
Possibly some acceleration method could be used
to actually get more accurate estimates.
A: Using the general relation
$$a_1\cdot\prod_{n=1}^\infty\frac{a_{n+1}}{a_n}=a_1+\sum_{n=1}^\infty \left(\frac{a_{n+1}}{a_n}-1\right)a_n$$ 
with the partial products
$$a_n\equiv p_{n-1}:=\prod_{k=1}^{n-1}\left(1+\dfrac{1}{k!}\right),$$
for which it happens that 
$$\frac{a_{n+1}}{a_n}-1=\frac{1}{n!},$$ 
I found the sum representation 
$$p=\sum_{n=0}^\infty\dfrac{1}{n!}\,{\prod_{k=1}^{n-1}}\left(1+\dfrac{1}{k!}\right)$$
The first few terms are $1,1,1,\tfrac{1}{2}$, for a total of 3.5. It is very fastly converging, as the monotonically increasing $a_n$ are, by definition, bounded by $p<4$. The fourth term is already $\dfrac{a_4}{4!}\approx 0.1\dots$
Here you can now also find a simple approximation scheme using the expansion of ${\rm e}^1$.
A: 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.
A: Just one observation $$\ln p=\sum_{n=1}^{\infty}\ln \left(1+\frac{1}{n!}\right)<\sum_{n=1}^{\infty}\frac{1}{n!}=e-1$$ Since $$\ln (1+x)-x< 0$$ for all $x> 0$. So, $$p<e^{e-1}\approx 5.5749\ldots$$
Additional Observation: A tighter lower and upper bound comes as below:
$$\ln p=\sum_{n\ge 1}\ln\left(1+\frac{1}{n!}\right)=\ln 2+\sum_{n\ge 2}\sum_{k\ge 1}\frac{(-1)^{k-1}}{k(n!)^k}\\ $$ Then using the inequality $$\left(\sum_{i}a_i^k\right)\le \left(\sum_{i}a_i\right)^k$$ for $a_i\ge 0$, we get (after some calculations, which is not very difficult)$$\ln 2+e-2+\frac{1}{2}\ln (1-(e-2)^2)<\ln p<\ln 2+\frac{1}{2}\ln\left(\frac{e-1}{3-e}\right)\\ \Rightarrow 2e^{e-2}\sqrt{4e-e^2-3}<p<2\sqrt{\frac{e-1}{3-e}}\\ \Rightarrow 2.8538\ldots <p< 4.9393\ldots$$
