How to compute $\prod_{n=1}^\infty\left(1+\frac{1}{n!}\right)$?

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)$$\ln 2+e-2+\frac{1}{2}\ln (1-(e-2)^2)<\ln p<\ln 2+\frac{1}{2}\ln\left(\frac{e-1}{3-e}\right)\\ \Rightarrow 2e^{e-2}\sqrt{4e-e^2-3}<p<2\sqrt{\frac{e-1}{3-e}}\\ \Rightarrow 2.8538\ldots <p< 4.9393\ldots$$

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.