Is $ \prod\limits_{k=0}^\infty \left(1 + \frac{1}{k!}\right) = \mathrm e^2 $? I was playing around and I came up with this product, which I believe to be equal to $\mathrm e^2$.
$$ \prod_{k=0}^\infty \left(1 + \frac{1}{k!}\right) \stackrel{?}{=} \mathrm e^2 $$
After calculating $1000$ terms of this product, I got approximately $7.36431$ (compare: $\mathrm e^2 \approx 7.38906 $, so convergence is very slow if existent).
I tried looking at some product definitions of $\rm e$, but none deal with the product I want.
I know that the product converges since $\sum_{k=0}^\infty 1/k!$ and $\sum_{k=0}^\infty 1/k!^2 $ converge as well.
 A: Fairly close, but as your computation shows, not close enough. The logarithm of the "rest" is not hard to estimate, and is far too close to $0$ to make up the gap.  
Remark: For small positive $x$, by an alternating series argument or otherwise, we have $\log(1+x)\lt x$. So the logarithm of the infinite product $\displaystyle\prod_{k=n}^\infty \left(1+\frac{1}{k!}\right)$ is less than $\displaystyle\sum_{k=n}^\infty \frac{1}{k!}$. 
By bounding the tail with a geometric series, we find this is $\lt \dfrac{n+1}{n\cdot n!}$.  The exponential of this, for large $n$, is much too close to $1$. A computation of the partial product to $n=8$ is more than  sufficient to show that we cannot have equality. 
The convergence of this infinite product is fast, in the same ballpark as the convergence of $\displaystyle \sum\dfrac{1}{k!}$.
A: Calculating the first 10 million terms of the product with a working precision of 2000 digits gives the result 
  7.36430827236725725637277250963105

At the moment I calculate less terms with a higher working precision but I would be surprised if the result would change a lot.
