Does it matter if e is irrational in order to solve to problem?

Heres another way to prove...

e - (1+1/1!+1/2!+...+1/n!) < 1/(n!n)

Well, e=1+1/1!+1/2!+..., so
e - (1+1/1!+1/2!+...+1/n!)=1/(n+1)!+1/(n+2)!+1/(n+3)!...
1/(n+1)!+1/(n+2)!+1/(n+3)!... is also equal to...
1/n!*1/(n+1)+1/n!*1/((n+1)(n+2))
in other words, if n=6 then we have...
1/6!*1/7+1/6!*1/(7*8)+1/6!*1/(7*8*9)....
And we need to prove this is less than
1/(n!n)
Which if n=6, then thats like 1/6!*1/6
A 1/6! can be factored out of both sides of the equation so, so far that we have...
1/7+1/(7*8)+1/(7*8*9)....<1/6
Now take a geometric series where the change is 1/7, it would look like..
1/7+1/(7*7)+1/(7*7*7)... we kind find the sum of all these with the common equation...
starting # / (1 - change)
(1/7)/(1-1/7)=1/6. Infact, for all sequences like this, they add up to 1/(n-1)
So, if the sequecnce with the 7's add up to 1/6, then the sequence 1/(7)+1/(7*8)+1/(7*8*9)... must be smaller than 1/6 because the numbers on the bottom get bigger than 7

e - (1+1/1!+1/2!+...+1/n!) < 1/(n!(n+1))
If this were the equation (notice the difference at the end), it would always be false because...
1/(n+1) will always be less than 1/(n+1)+some other stuff...

If it looked like this....
e - (1+1/1!+1/2!+...+1/n!) < 1/(n!(n+.5))
I think it would start to depend upon what n was.