Double natural constants?

This is same as:

$\sum _{ m=1 }^{ \infty }{ \sum _{ k=m+1 }^{ \infty }{ \frac { 1 }{ k! } } } \\ =\sum _{ k=1 }^{ \infty }{ \frac { k-1 }{ k! } } \\ =\sum _{ k=1 }^{ \infty }{ \frac { k }{ k! } } -\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ k! } } \\ =\frac { 1 }{ 0! } +\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ (k-1)! } } -\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ k! } } \\ =1+\sum _{ j=1 }^{ \infty }{ \frac { 1 }{ j! } } -\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ k! } }$

Now we note that $\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ k! } }$ is absolutely convergent, so the rearrangement of terms is valid, therefore $\sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (m+n)! } } } =1$

Expanding the series for a few terms, we can see that it is basically $\sum_{i=2}^\infty \frac{(n-1)}{n!}$ . From the series expansion of the exponential function, we can see that the above series matches the series expansion of the differential of $\frac{{e}^{x}}{x}-\frac{1}{x}=\frac{{e}^{x}(x-1)+1}{{x}^{2}}$ for when x=1. Putting x=1 in the function above, we can see that the answer is $\boxed1$

Telescoping the series we terminate it into 1/n! - 1/(n+1)! Where n ranges from 1 to infinite. Hence after solving limits we get answer=1