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.